Class 12 Maths Chapter 7 Integrals Exercise 7.5 NCERT Solutions

Exercise 7.5: Integration by Partial Fractions

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• Ex. 7.1


• Ex. 7.2


• Ex. 7.2


• Ex. 7.3


• Ex. 7.4


• Ex. 7.5


• Ex. 7.6


• Ex. 7.7


• Ex. 7.8


• Ex. 7.9


• Ex. 7.10


• Ex. 7.11


• Miscellaneous Exercise

Question 1

Integrate the rational function: \(\frac{x}{(x+1)(x+2)}\)

Solution:

Let \(\frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}\)

\(\Rightarrow x = A(x+2) + B(x+1) \quad \dots(1)\)

Putting \(x = -1\) in (1):
\(-1 = A(1) \Rightarrow A = -1\)

Putting \(x = -2\) in (1):
\(-2 = B(-1) \Rightarrow B = 2\)

Therefore,
\(\int \frac{x}{(x+1)(x+2)} dx = \int \left( \frac{-1}{x+1} + \frac{2}{x+2} \right) dx\)
\(= -\log|x+1| + 2\log|x+2| + C\)
\(= \log(x+2)^2 - \log|x+1| + C\)

Question 2

Integrate the rational function: \(\frac{1}{x^2 - 9}\)

Solution:

The expression can be factored as \(\frac{1}{(x-3)(x+3)}\).

Let \(\frac{1}{(x-3)(x+3)} = \frac{A}{x-3} + \frac{B}{x+3}\)

\(\Rightarrow 1 = A(x+3) + B(x-3) \quad \dots(1)\)

Putting \(x = 3\) in (1):
\(1 = 6A \Rightarrow A = \frac{1}{6}\)

Putting \(x = -3\) in (1):
\(1 = -6B \Rightarrow B = -\frac{1}{6}\)

Therefore,
\(\int \frac{dx}{x^2 - 9} = \frac{1}{6} \int \frac{dx}{x-3} - \frac{1}{6} \int \frac{dx}{x+3}\)
\(= \frac{1}{6} \log|x-3| - \frac{1}{6} \log|x+3| + C\)
\(= \frac{1}{6} \log \left| \frac{x-3}{x+3} \right| + C\)

Question 3

Integrate the rational function: \(\frac{3x-1}{(x-1)(x-2)(x-3)}\)

Solution:

Let \(\frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}\)

\(\Rightarrow 3x-1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)\)

Putting \(x=1\): \(2 = A(-1)(-2) \Rightarrow 2A = 2 \Rightarrow A = 1\)

Putting \(x=2\): \(5 = B(1)(-1) \Rightarrow -B = 5 \Rightarrow B = -5\)

Putting \(x=3\): \(8 = C(2)(1) \Rightarrow 2C = 8 \Rightarrow C = 4\)

Thus,
\(\int \frac{3x-1}{(x-1)(x-2)(x-3)} dx = \int \frac{1}{x-1} dx - 5 \int \frac{1}{x-2} dx + 4 \int \frac{1}{x-3} dx\)
\(= \log|x-1| - 5\log|x-2| + 4\log|x-3| + C\)

Question 4

Integrate the rational function: \(\frac{x}{(x-1)(x-2)(x-3)}\)

Solution:

Let \(\frac{x}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}\)

\(\Rightarrow x = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)\)

Putting \(x=1\): \(1 = 2A \Rightarrow A = \frac{1}{2}\)

Putting \(x=2\): \(2 = -B \Rightarrow B = -2\)

Putting \(x=3\): \(3 = 2C \Rightarrow C = \frac{3}{2}\)

Therefore,
\(\int \frac{x}{(x-1)(x-2)(x-3)} dx = \frac{1}{2}\log|x-1| - 2\log|x-2| + \frac{3}{2}\log|x-3| + C\)

Question 5

Integrate the rational function: \(\frac{2x}{x^2+3x+2}\)

Solution:

Denominator \(x^2+3x+2 = (x+1)(x+2)\).

Let \(\frac{2x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}\)

\(\Rightarrow 2x = A(x+2) + B(x+1)\)

Putting \(x=-1\): \(-2 = A \Rightarrow A = -2\)

Putting \(x=-2\): \(-4 = -B \Rightarrow B = 4\)

Thus,
\(\int \frac{2x}{x^2+3x+2} dx = -2\log|x+1| + 4\log|x+2| + C\)

Question 6

Integrate the rational function: \(\frac{1-x^2}{x(1-2x)}\)

Solution:

The integrand is \(\frac{1-x^2}{x-2x^2}\). Since the degree of the numerator equals the degree of the denominator, we divide first.

\(\frac{1-x^2}{-2x^2+x} = \frac{1}{2} + \frac{1 - \frac{x}{2}}{x(1-2x)} = \frac{1}{2} + \frac{2-x}{2x(1-2x)}\)

Let \(\frac{2-x}{x(1-2x)} = \frac{A}{x} + \frac{B}{1-2x}\)

\(\Rightarrow 2-x = A(1-2x) + Bx\)

Putting \(x=0\): \(A=2\)

Putting \(x=\frac{1}{2}\): \(\frac{3}{2} = \frac{B}{2} \Rightarrow B=3\)

Thus, Integrand = \(\frac{1}{2} + \frac{1}{2} \left( \frac{2}{x} + \frac{3}{1-2x} \right)\)

\(\int \left( \frac{1}{2} + \frac{1}{x} + \frac{3}{2(1-2x)} \right) dx\)
\(= \frac{x}{2} + \log|x| + \frac{3}{2} \cdot \frac{\log|1-2x|}{-2} + C\)
\(= \frac{x}{2} + \log|x| - \frac{3}{4}\log|1-2x| + C\)

Question 7

Integrate the rational function: \(\frac{x}{(x^2+1)(x-1)}\)

Solution:

Let \(\frac{x}{(x^2+1)(x-1)} = \frac{Ax+B}{x^2+1} + \frac{C}{x-1}\)

\(\Rightarrow x = (Ax+B)(x-1) + C(x^2+1)\)

Putting \(x=1\): \(1 = 2C \Rightarrow C = \frac{1}{2}\)

Comparing coefficients of \(x^2\): \(A+C=0 \Rightarrow A = -\frac{1}{2}\)

Comparing constant terms: \(-B+C=0 \Rightarrow B = \frac{1}{2}\)

Integral = \(\int \left( \frac{-\frac{1}{2}x + \frac{1}{2}}{x^2+1} + \frac{\frac{1}{2}}{x-1} \right) dx\)

\(= -\frac{1}{2}\int \frac{x}{x^2+1} dx + \frac{1}{2}\int \frac{1}{x^2+1} dx + \frac{1}{2}\int \frac{1}{x-1} dx\)

\(= -\frac{1}{4}\log(x^2+1) + \frac{1}{2}\tan^{-1}x + \frac{1}{2}\log|x-1| + C\)

Question 8

Integrate the rational function: \(\frac{x}{(x-1)^2(x+2)}\)

Solution:

Let \(\frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}\)

\(\Rightarrow x = A(x-1)(x+2) + B(x+2) + C(x-1)^2\)

Putting \(x=1\): \(1 = 3B \Rightarrow B = \frac{1}{3}\)

Putting \(x=-2\): \(-2 = 9C \Rightarrow C = -\frac{2}{9}\)

Comparing coeff of \(x^2\): \(A+C=0 \Rightarrow A = \frac{2}{9}\)

Integral = \(\frac{2}{9}\log|x-1| + \frac{1}{3} \int (x-1)^{-2} dx - \frac{2}{9}\log|x+2| + C\)

\(= \frac{2}{9}\log \left| \frac{x-1}{x+2} \right| - \frac{1}{3(x-1)} + C\)

Question 9

Integrate the rational function: \(\frac{3x+5}{x^3-x^2-x+1}\)

Solution:

\(x^3-x^2-x+1 = x^2(x-1) - 1(x-1) = (x^2-1)(x-1) = (x+1)(x-1)^2\)

Let \(\frac{3x+5}{(x+1)(x-1)^2} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{(x-1)^2}\)

\(\Rightarrow 3x+5 = A(x-1)^2 + B(x+1)(x-1) + C(x+1)\)

Putting \(x=1\): \(8 = 2C \Rightarrow C=4\)

Putting \(x=-1\): \(2 = 4A \Rightarrow A=\frac{1}{2}\)

Coeff of \(x^2\): \(A+B=0 \Rightarrow B=-\frac{1}{2}\)

Integral = \(\frac{1}{2}\log|x+1| - \frac{1}{2}\log|x-1| - \frac{4}{x-1} + C\)

\(= \frac{1}{2} \log \left| \frac{x+1}{x-1} \right| - \frac{4}{x-1} + C\)

Question 10

Integrate the rational function: \(\frac{2x-3}{(x^2-1)(2x+3)}\)

Solution:

Factor: \((x-1)(x+1)(2x+3)\)

Let \(\frac{2x-3}{(x-1)(x+1)(2x+3)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{2x+3}\)

\(x=1 \Rightarrow -1 = A(2)(5) \Rightarrow A = -\frac{1}{10}\)

\(x=-1 \Rightarrow -5 = B(-2)(1) \Rightarrow B = \frac{5}{2}\)

\(x=-\frac{3}{2} \Rightarrow -6 = C(-\frac{5}{2})(-\frac{1}{2}) \Rightarrow C = -\frac{24}{5}\)

Integral = \(-\frac{1}{10}\log|x-1| + \frac{5}{2}\log|x+1| - \frac{24}{5} \frac{\log|2x+3|}{2} + C\)

\(= \frac{5}{2}\log|x+1| - \frac{1}{10}\log|x-1| - \frac{12}{5}\log|2x+3| + C\)

Question 11

Integrate the rational function: \(\frac{5x}{(x+1)(x^2-4)}\)

Solution:

Factors: \((x+1)(x-2)(x+2)\). Let \(\frac{5x}{(x+1)(x-2)(x+2)} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+2}\)

\(x=-1 \Rightarrow -5 = A(-3)(1) \Rightarrow A = \frac{5}{3}\)

\(x=2 \Rightarrow 10 = B(3)(4) \Rightarrow B = \frac{5}{6}\)

\(x=-2 \Rightarrow -10 = C(-1)(-4) \Rightarrow C = -\frac{5}{2}\)

Integral = \(\frac{5}{3}\log|x+1| + \frac{5}{6}\log|x-2| - \frac{5}{2}\log|x+2| + C\)

Question 12

Integrate the rational function: \(\frac{x^3+x+1}{x^2-1}\)

Solution:

Divide numerator by denominator: \(x^3+x+1 = x(x^2-1) + 2x+1\).

Integrand = \(x + \frac{2x+1}{x^2-1} = x + \frac{2x+1}{(x-1)(x+1)}\)

Let \(\frac{2x+1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}\)

\(x=1 \Rightarrow 3 = 2A \Rightarrow A = \frac{3}{2}\)

\(x=-1 \Rightarrow -1 = -2B \Rightarrow B = \frac{1}{2}\)

Integral = \(\frac{x^2}{2} + \frac{3}{2}\log|x-1| + \frac{1}{2}\log|x+1| + C\)

Question 13

Integrate the rational function: \(\frac{2}{(1-x)(1+x^2)}\)

Solution:

Let \(\frac{2}{(1-x)(1+x^2)} = \frac{A}{1-x} + \frac{Bx+C}{1+x^2}\)

\(2 = A(1+x^2) + (Bx+C)(1-x)\)

\(x=1 \Rightarrow 2 = 2A \Rightarrow A=1\)

Coeff \(x^2\): \(A-B=0 \Rightarrow B=1\)

Constant: \(A+C=2 \Rightarrow C=1\)

Integral = \(\int \frac{1}{1-x} dx + \int \frac{x}{1+x^2} dx + \int \frac{1}{1+x^2} dx\)

\(= -\log|1-x| + \frac{1}{2}\log(1+x^2) + \tan^{-1}x + C\)

Question 14

Integrate the rational function: \(\frac{3x-1}{(x+2)^2}\)

Solution:

Let \(\frac{3x-1}{(x+2)^2} = \frac{A}{x+2} + \frac{B}{(x+2)^2}\)

\(3x-1 = A(x+2) + B\)

\(A=3\), \(2A+B=-1 \Rightarrow 6+B=-1 \Rightarrow B=-7\)

Integral = \(3\log|x+2| + \frac{-7}{(x+2)} \cdot (-1) \text{ (Oops, integral of } u^{-2} \text{ is } -u^{-1})\)

\(= 3\log|x+2| + \frac{7}{x+2} + C\)

Question 15

Integrate the rational function: \(\frac{1}{x^4-1}\)

Solution:

\(\frac{1}{x^4-1} = \frac{1}{(x^2-1)(x^2+1)} = \frac{1}{2} \left[ \frac{1}{x^2-1} - \frac{1}{x^2+1} \right]\)

\(= \frac{1}{2(x-1)(x+1)} - \frac{1}{2(x^2+1)}\)

\(= \frac{1}{2} \cdot \frac{1}{2} \left( \frac{1}{x-1} - \frac{1}{x+1} \right) - \frac{1}{2(x^2+1)}\)

\(= \frac{1}{4(x-1)} - \frac{1}{4(x+1)} - \frac{1}{2(x^2+1)}\)

Integral = \(\frac{1}{4}\log|x-1| - \frac{1}{4}\log|x+1| - \frac{1}{2}\tan^{-1}x + C\)

\(= \frac{1}{4}\log \left| \frac{x-1}{x+1} \right| - \frac{1}{2}\tan^{-1}x + C\)

Question 16

Integrate \(\frac{1}{x(x^n + 1)}\)

Solution:

Multiply numerator and denominator by \(x^{n-1}\): \(\int \frac{x^{n-1}}{x^n(x^n+1)} dx\).

Put \(x^n = t \Rightarrow nx^{n-1}dx = dt\).

Integral = \(\frac{1}{n} \int \frac{dt}{t(t+1)} = \frac{1}{n} \int \left( \frac{1}{t} - \frac{1}{t+1} \right) dt\)

\(= \frac{1}{n} \left( \log|t| - \log|t+1| \right) + C\)

\(= \frac{1}{n} \log \left| \frac{x^n}{x^n+1} \right| + C\)

Question 17

Integrate \(\frac{\cos x}{(1-\sin x)(2-\sin x)}\)

Solution:

Put \(\sin x = t \Rightarrow \cos x dx = dt\).

Integrand becomes \(\frac{1}{(1-t)(2-t)}\).

Partial fractions: \(\frac{1}{(1-t)(2-t)} = \frac{1}{1-t} - \frac{1}{2-t}\) (Check: \(\frac{(2-t)-(1-t)}{(1-t)(2-t)} = \frac{1}{\dots}\))

Integral = \(\int \frac{dt}{1-t} - \int \frac{dt}{2-t}\)

\(= -\log|1-t| - (-\log|2-t|) + C\)

\(= \log|2-t| - \log|1-t| + C = \log \left| \frac{2-\sin x}{1-\sin x} \right| + C\)

Question 18

Integrate \(\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}\)

Solution:

Let \(x^2 = y\) for partial fractions. \(\frac{(y+1)(y+2)}{(y+3)(y+4)} = \frac{y^2+3y+2}{y^2+7y+12} = 1 - \frac{4y+10}{(y+3)(y+4)}\).

\(\frac{4y+10}{(y+3)(y+4)} = \frac{A}{y+3} + \frac{B}{y+4}\)

\(y=-3 \Rightarrow -2 = A\). \(y=-4 \Rightarrow -6 = -B \Rightarrow B=6\).

Integrand = \(1 - \left( \frac{-2}{x^2+3} + \frac{6}{x^2+4} \right) = 1 + \frac{2}{x^2+3} - \frac{6}{x^2+4}\)

Integral = \(x + \frac{2}{\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}} - 3\tan^{-1}\frac{x}{2} + C\)

Question 19

Integrate \(\frac{2x}{(x^2+1)(x^2+3)}\)

Solution:

Put \(x^2 = t \Rightarrow 2x dx = dt\).

Integrand = \(\frac{1}{(t+1)(t+3)} = \frac{1}{2} \left( \frac{1}{t+1} - \frac{1}{t+3} \right)\).

Integral = \(\frac{1}{2} \left( \log|t+1| - \log|t+3| \right) + C\)

\(= \frac{1}{2} \log \left( \frac{x^2+1}{x^2+3} \right) + C\)

Question 20

Integrate \(\frac{1}{x(x^4-1)}\)

Solution:

Use result from Q16 with \(n=4\), but note sign change if using \(t=x^4\).

\(\int \frac{x^3 dx}{x^4(x^4-1)}\). Let \(x^4=t\). \(\frac{1}{4}\int \frac{dt}{t(t-1)}\).

\(\frac{1}{4} \int \left( \frac{1}{t-1} - \frac{1}{t} \right) dt = \frac{1}{4} \log \left| \frac{t-1}{t} \right| + C\)

\(= \frac{1}{4} \log \left| \frac{x^4-1}{x^4} \right| + C\)

Question 21

Integrate \(\frac{1}{e^x - 1}\)

Solution:

Put \(e^x = t \Rightarrow e^x dx = dt \Rightarrow dx = \frac{dt}{t}\).

Integral = \(\int \frac{dt}{t(t-1)}\). Same as Q20 partial fractions.

\(= \int \left( \frac{1}{t-1} - \frac{1}{t} \right) dt = \log|t-1| - \log|t| + C\)

\(= \log \left| \frac{e^x-1}{e^x} \right| + C = \log|1 - e^{-x}| + C\)

Question 22

\(\int \frac{x dx}{(x-1)(x-2)}\) equals

A. \(\log| (x-1)^2/(x-2)|+C\) B. \(\log| (x-2)^2/(x-1)|+C\) C. \(\log| (x-1)/(x-2)^2|+C\) D. \(\log|(x-1)(x-2)|+C\)

Solution:

\(\frac{x}{(x-1)(x-2)} = \frac{-1}{x-1} + \frac{2}{x-2}\).

Integral = \(-\log|x-1| + 2\log|x-2| + C = \log \left| \frac{(x-2)^2}{x-1} \right| + C\).

Correct Answer: B

Question 23

\(\int \frac{dx}{x(x^2+1)}\) equals

A. \(\log|x| - \frac{1}{2} \log (x^2 + 1) + C\) B. \(\log|x| + \frac{1}{2} \log (x^2 + 1) + C\) C. \(-\log|x| + \frac{1}{2} \log (x^2 + 1) + C\) D. \(\frac{1}{2} \log|x| + \log (x^2 + 1) + C\)

Solution:

\(\frac{1}{x(x^2+1)} = \frac{1}{x} - \frac{x}{x^2+1}\).

Integral = \(\log|x| - \frac{1}{2}\log(x^2+1) + C\).

Correct Answer: A

NCERT Solutions Class 12 Maths Chapter 7 Integrals Exercise 7.3

Download Ex. 7.3 Solution PDF

• Ex. 7.1


• Ex. 7.2


• Ex. 7.2


• Ex. 7.3


• Ex. 7.4


• Ex. 7.5


• Ex. 7.6


• Ex. 7.7


• Ex. 7.8


• Ex. 7.9


• Ex. 7.10


• Ex. 7.11


• Miscellaneous Exercise

Integrals Class 12 Mathematics Exercise 7.3

Find the integrals of the functions in Exercises 1 to 22.

Question 1.

\(\sin^2 (2x+5)\)

Answer:

We know the identity \(\sin^2 A = \frac{1 - \cos 2A}{2}\).
Therefore, \[ \int \sin^2(2x+5) \, dx = \int \frac{1 - \cos 2(2x+5)}{2} \, dx \] \[ = \int \frac{1 - \cos(4x+10)}{2} \, dx \] \[ = \frac{1}{2} \int 1 \, dx - \frac{1}{2} \int \cos(4x+10) \, dx \] \[ = \frac{1}{2}x - \frac{1}{2} \cdot \frac{\sin(4x+10)}{4} + C \] \[ = \frac{x}{2} - \frac{1}{8}\sin(4x+10) + C \]

Question 2.

\(\sin 3x \cos 4x\)

Answer:

Using the identity \(2\sin A \cos B = \sin(A+B) + \sin(A-B)\): \[ \int \sin 3x \cos 4x \, dx = \frac{1}{2} \int 2\sin 3x \cos 4x \, dx \] \[ = \frac{1}{2} \int [\sin(3x+4x) + \sin(3x-4x)] \, dx \] \[ = \frac{1}{2} \int [\sin 7x + \sin(-x)] \, dx \] \[ = \frac{1}{2} \int (\sin 7x - \sin x) \, dx \] \[ = \frac{1}{2} \left[ \frac{-\cos 7x}{7} - (-\cos x) \right] + C \] \[ = \frac{-\cos 7x}{14} + \frac{\cos x}{2} + C \]

Question 3.

\(\cos 2x \cos 4x \cos 6x\)

Answer:

Let \(I = \int \cos 2x \cos 4x \cos 6x \, dx\).
Using \(2\cos A \cos B = \cos(A+B) + \cos(A-B)\): \[ I = \int \cos 2x \cdot \frac{1}{2}(2\cos 4x \cos 6x) \, dx \] \[ = \frac{1}{2} \int \cos 2x [\cos(4x+6x) + \cos(4x-6x)] \, dx \] \[ = \frac{1}{2} \int \cos 2x (\cos 10x + \cos 2x) \, dx \] \[ = \frac{1}{2} \int (\cos 2x \cos 10x + \cos^2 2x) \, dx \] Again applying product identity and \(\cos^2 A = \frac{1+\cos 2A}{2}\): \[ = \frac{1}{2} \int \left[ \frac{1}{2}(\cos 12x + \cos 8x) + \frac{1+\cos 4x}{2} \right] \, dx \] \[ = \frac{1}{4} \int (\cos 12x + \cos 8x + 1 + \cos 4x) \, dx \] \[ = \frac{1}{4} \left[ \frac{\sin 12x}{12} + \frac{\sin 8x}{8} + x + \frac{\sin 4x}{4} \right] + C \]

Question 4.

\(\sin^3(2x+1)\)

Answer:

Let \(I = \int \sin^3(2x+1) \, dx\).
We can write \(\sin^3 \theta = \sin^2 \theta \cdot \sin \theta = (1-\cos^2 \theta)\sin \theta\). \[ I = \int (1-\cos^2(2x+1))\sin(2x+1) \, dx \] Let \(\cos(2x+1) = t\). Then \(-2\sin(2x+1) \, dx = dt \implies \sin(2x+1) \, dx = \frac{-dt}{2}\). \[ I = \frac{-1}{2} \int (1-t^2) \, dt \] \[ = \frac{-1}{2} \left[ t - \frac{t^3}{3} \right] + C \] Substituting \(t = \cos(2x+1)\): \[ = \frac{-1}{2} \cos(2x+1) + \frac{1}{6} \cos^3(2x+1) + C \]

Question 5.

\(\sin^3 x \cos^3 x\)

Answer:

Let \(I = \int \sin^3 x \cos^3 x \, dx\). \[ I = \int \sin x \cdot \sin^2 x \cdot \cos^3 x \, dx \] \[ I = \int \sin x (1-\cos^2 x) \cos^3 x \, dx \] Let \(\cos x = t \implies -\sin x \, dx = dt\). \[ I = -\int (1-t^2)t^3 \, dt = -\int (t^3 - t^5) \, dt \] \[ = -\left( \frac{t^4}{4} - \frac{t^6}{6} \right) + C \] \[ = \frac{\cos^6 x}{6} - \frac{\cos^4 x}{4} + C \]

Question 6.

\(\sin x \sin 2x \sin 3x\)

Answer:

\[ I = \int \sin x \sin 2x \sin 3x \, dx \] Using \(2\sin A \sin B = \cos(A-B) - \cos(A+B)\): \[ = \frac{1}{2} \int \sin 3x (2\sin x \sin 2x) \, dx \] \[ = \frac{1}{2} \int \sin 3x (\cos x - \cos 3x) \, dx \] \[ = \frac{1}{2} \int (\sin 3x \cos x - \sin 3x \cos 3x) \, dx \] Multiply and divide by 2 again: \[ = \frac{1}{4} \int (2\sin 3x \cos x - 2\sin 3x \cos 3x) \, dx \] \[ = \frac{1}{4} \int [(\sin 4x + \sin 2x) - \sin 6x] \, dx \] \[ = \frac{1}{4} \left[ -\frac{\cos 4x}{4} - \frac{\cos 2x}{2} - \left(-\frac{\cos 6x}{6}\right) \right] + C \] \[ = \frac{1}{4} \left[ \frac{\cos 6x}{6} - \frac{\cos 4x}{4} - \frac{\cos 2x}{2} \right] + C \]

Question 7.

\(\sin 4x \sin 8x\)

Answer:

Using \(2\sin A \sin B = \cos(A-B) - \cos(A+B)\): \[ \int \sin 4x \sin 8x \, dx = \frac{1}{2} \int 2\sin 4x \sin 8x \, dx \] \[ = \frac{1}{2} \int [\cos(4x-8x) - \cos(4x+8x)] \, dx \] \[ = \frac{1}{2} \int (\cos 4x - \cos 12x) \, dx \] \[ = \frac{1}{2} \left[ \frac{\sin 4x}{4} - \frac{\sin 12x}{12} \right] + C \]

Question 8.

\(\frac{1-\cos x}{1+\cos x}\)

Answer:

Using half-angle identities \(1-\cos x = 2\sin^2(x/2)\) and \(1+\cos x = 2\cos^2(x/2)\): \[ \int \frac{1-\cos x}{1+\cos x} \, dx = \int \frac{2\sin^2(x/2)}{2\cos^2(x/2)} \, dx \] \[ = \int \tan^2(x/2) \, dx \] Using \(\tan^2 \theta = \sec^2 \theta - 1\): \[ = \int (\sec^2(x/2) - 1) \, dx \] \[ = \frac{\tan(x/2)}{1/2} - x + C \] \[ = 2\tan(x/2) - x + C \]

Question 9.

\(\frac{\cos x}{1+\cos x}\)

Answer:

\[ \int \frac{\cos x}{1+\cos x} \, dx = \int \frac{1+\cos x - 1}{1+\cos x} \, dx \] \[ = \int \left( 1 - \frac{1}{1+\cos x} \right) \, dx \] \[ = \int \left( 1 - \frac{1}{2\cos^2(x/2)} \right) \, dx \] \[ = \int \left( 1 - \frac{1}{2}\sec^2(x/2) \right) \, dx \] \[ = x - \frac{1}{2} \cdot \frac{\tan(x/2)}{1/2} + C \] \[ = x - \tan(x/2) + C \]

Question 10.

\(\sin^4 x\)

Answer:

\[ \sin^4 x = (\sin^2 x)^2 = \left( \frac{1-\cos 2x}{2} \right)^2 \] \[ = \frac{1}{4} (1 - 2\cos 2x + \cos^2 2x) \] \[ = \frac{1}{4} \left[ 1 - 2\cos 2x + \frac{1+\cos 4x}{2} \right] \] \[ = \frac{1}{4} \left[ 1 - 2\cos 2x + \frac{1}{2} + \frac{1}{2}\cos 4x \right] \] \[ = \frac{1}{4} \left[ \frac{3}{2} - 2\cos 2x + \frac{1}{2}\cos 4x \right] \] Integrating term by term: \[ \int \sin^4 x \, dx = \frac{1}{4} \left[ \frac{3}{2}x - \sin 2x + \frac{\sin 4x}{8} \right] + C \] \[ = \frac{3x}{8} - \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + C \]

Question 11.

\(\cos^4 2x\)

Answer:

\[ \cos^4 2x = (\cos^2 2x)^2 = \left( \frac{1+\cos 4x}{2} \right)^2 \] \[ = \frac{1}{4} (1 + 2\cos 4x + \cos^2 4x) \] \[ = \frac{1}{4} \left[ 1 + 2\cos 4x + \frac{1+\cos 8x}{2} \right] \] \[ = \frac{1}{4} \left[ \frac{3}{2} + 2\cos 4x + \frac{1}{2}\cos 8x \right] \] Integrating: \[ \int \cos^4 2x \, dx = \frac{1}{4} \left[ \frac{3}{2}x + \frac{2\sin 4x}{4} + \frac{\sin 8x}{16} \right] + C \] \[ = \frac{3x}{8} + \frac{1}{8}\sin 4x + \frac{1}{64}\sin 8x + C \]

Question 12.

\(\frac{\sin^2 x}{1+\cos x}\)

Answer:

\[ \int \frac{\sin^2 x}{1+\cos x} \, dx = \int \frac{1-\cos^2 x}{1+\cos x} \, dx \] \[ = \int \frac{(1-\cos x)(1+\cos x)}{1+\cos x} \, dx \] \[ = \int (1-\cos x) \, dx \] \[ = x - \sin x + C \]

Question 13.

\(\frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}\)

Answer:

Using \(\cos 2\theta = 2\cos^2 \theta - 1\): \[ \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} = \frac{(2\cos^2 x - 1) - (2\cos^2 \alpha - 1)}{\cos x - \cos \alpha} \] \[ = \frac{2(\cos^2 x - \cos^2 \alpha)}{\cos x - \cos \alpha} \] \[ = \frac{2(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} \] \[ = 2(\cos x + \cos \alpha) \] Now integrating: \[ \int 2(\cos x + \cos \alpha) \, dx = 2(\sin x + x\cos \alpha) + C \]

Question 14.

\(\frac{\cos x - \sin x}{1+\sin 2x}\)

Answer:

We know that \(1+\sin 2x = \sin^2 x + \cos^2 x + 2\sin x \cos x = (\sin x + \cos x)^2\). \[ I = \int \frac{\cos x - \sin x}{(\sin x + \cos x)^2} \, dx \] Let \(\sin x + \cos x = t\). Then \((\cos x - \sin x) \, dx = dt\). \[ I = \int \frac{dt}{t^2} = \int t^{-2} \, dt = -t^{-1} + C \] \[ = \frac{-1}{\sin x + \cos x} + C \]

Question 15.

\(\tan^3 2x \sec 2x\)

Answer:

\[ \int \tan^3 2x \sec 2x \, dx = \int \tan^2 2x \cdot \sec 2x \tan 2x \, dx \] \[ = \int (\sec^2 2x - 1) \sec 2x \tan 2x \, dx \] Let \(\sec 2x = t\). Then \(2\sec 2x \tan 2x \, dx = dt \implies \sec 2x \tan 2x \, dx = \frac{dt}{2}\). \[ I = \frac{1}{2} \int (t^2 - 1) \, dt \] \[ = \frac{1}{2} \left[ \frac{t^3}{3} - t \right] + C \] \[ = \frac{1}{6}\sec^3 2x - \frac{1}{2}\sec 2x + C \]

Question 16.

\(\tan^4 x\)

Answer:

\[ \int \tan^4 x \, dx = \int \tan^2 x \cdot \tan^2 x \, dx \] \[ = \int \tan^2 x (\sec^2 x - 1) \, dx \] \[ = \int \tan^2 x \sec^2 x \, dx - \int \tan^2 x \, dx \] \[ = \int \tan^2 x \sec^2 x \, dx - \int (\sec^2 x - 1) \, dx \] For the first part, let \(\tan x = t \implies \sec^2 x \, dx = dt\). \[ \int t^2 \, dt - (\tan x - x) \] \[ = \frac{t^3}{3} - \tan x + x + C \] \[ = \frac{1}{3}\tan^3 x - \tan x + x + C \]

Question 17.

\(\frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x}\)

Answer:

Split the terms: \[ \int \frac{\sin^3 x}{\sin^2 x \cos^2 x} \, dx + \int \frac{\cos^3 x}{\sin^2 x \cos^2 x} \, dx \] \[ = \int \frac{\sin x}{\cos^2 x} \, dx + \int \frac{\cos x}{\sin^2 x} \, dx \] \[ = \int \tan x \sec x \, dx + \int \cot x \csc x \, dx \] \[ = \sec x - \csc x + C \]

Question 18.

\(\frac{\cos 2x + 2\sin^2 x}{\cos^2 x}\)

Answer:

Using \(\cos 2x = 1 - 2\sin^2 x\): \[ \text{Numerator} = (1 - 2\sin^2 x) + 2\sin^2 x = 1 \] \[ \therefore I = \int \frac{1}{\cos^2 x} \, dx = \int \sec^2 x \, dx \] \[ = \tan x + C \]

Question 19.

\(\frac{1}{\sin x \cos^3 x}\)

Answer:

Substitute \(1 = \sin^2 x + \cos^2 x\): \[ I = \int \frac{\sin^2 x + \cos^2 x}{\sin x \cos^3 x} \, dx \] \[ = \int \frac{\sin^2 x}{\sin x \cos^3 x} \, dx + \int \frac{\cos^2 x}{\sin x \cos^3 x} \, dx \] \[ = \int \frac{\sin x}{\cos^3 x} \, dx + \int \frac{1}{\sin x \cos x} \, dx \] \[ = \int \tan x \sec^2 x \, dx + \int \frac{1}{\sin x \cos x} \cdot \frac{\cos x}{\cos x} \, dx \] \[ = \int \tan x \sec^2 x \, dx + \int \frac{\sec^2 x}{\tan x} \, dx \] Let \(\tan x = t \implies \sec^2 x \, dx = dt\). \[ I = \int t \, dt + \int \frac{1}{t} \, dt \] \[ = \frac{t^2}{2} + \log|t| + C \] \[ = \frac{1}{2}\tan^2 x + \log|\tan x| + C \]

Question 20.

\(\frac{\cos 2x}{(\cos x + \sin x)^2}\)

Answer:

Use \(\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)\): \[ I = \int \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\cos x + \sin x)^2} \, dx \] \[ = \int \frac{\cos x - \sin x}{\cos x + \sin x} \, dx \] Let \(\cos x + \sin x = t\). Then \((-\sin x + \cos x) \, dx = dt\). \[ I = \int \frac{dt}{t} = \log|t| + C \] \[ = \log|\cos x + \sin x| + C \]

Question 21.

\(\sin^{-1}(\cos x)\)

Answer:

We know \(\cos x = \sin(\frac{\pi}{2} - x)\). \[ I = \int \sin^{-1}\left( \sin\left(\frac{\pi}{2} - x\right) \right) \, dx \] \[ = \int \left( \frac{\pi}{2} - x \right) \, dx \] \[ = \frac{\pi}{2}x - \frac{x^2}{2} + C \]

Question 22.

\(\frac{1}{\cos(x-a)\cos(x-b)}\)

Answer:

Multiply and divide by \(\sin(a-b)\): \[ I = \frac{1}{\sin(a-b)} \int \frac{\sin((x-b)-(x-a))}{\cos(x-a)\cos(x-b)} \, dx \] Expand numerator using \(\sin(A-B)\): \[ = \frac{1}{\sin(a-b)} \int \frac{\sin(x-b)\cos(x-a) - \cos(x-b)\sin(x-a)}{\cos(x-a)\cos(x-b)} \, dx \] \[ = \frac{1}{\sin(a-b)} \int [\tan(x-b) - \tan(x-a)] \, dx \] \[ = \frac{1}{\sin(a-b)} [-\log|\cos(x-b)| - (-\log|\cos(x-a)|)] + C \] \[ = \frac{1}{\sin(a-b)} \log \left| \frac{\cos(x-a)}{\cos(x-b)} \right| + C \]

Choose the correct answer in Exercises 23 and 24.

Question 23.

\(\int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} \, dx\) is equal to

A. \(\tan x + \cot x + C\)

B. \(\tan x + \text{cosec } x + C\)

C. \(-\tan x + \cot x + C\)

D. \(\tan x + \sec x + C\)

Answer: A

\[ \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} - \frac{\cos^2 x}{\sin^2 x \cos^2 x} \right) \, dx \] \[ = \int (\sec^2 x - \text{cosec}^2 x) \, dx \] \[ = \tan x - (-\cot x) + C \] \[ = \tan x + \cot x + C \]

Question 24.

\(\int \frac{e^x(1+x)}{\cos^2(e^x x)} \, dx\) equals

A. \(-\cot(e^x x) + C\)

B. \(\tan(x e^x) + C\)

C. \(\tan(e^x) + C\)

D. \(\cot(e^x) + C\)

Answer: B

Let \(x e^x = t\).
Differentiating using product rule: \((e^x \cdot 1 + x \cdot e^x) \, dx = dt\).
\(e^x(1+x) \, dx = dt\).
Substituting into integral: \[ \int \frac{dt}{\cos^2 t} = \int \sec^2 t \, dt \] \[ = \tan t + C \] \[ = \tan(x e^x) + C \]

NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10 Integrals

Exercise 7.10: Definite Integrals by Substitution

Detailed step-by-step solutions for NCERT Class 12 Mathematics, Chapter 7, Exercise 7.10. These problems focus on evaluating definite integrals using the method of substitution.

⬇ Download Ex. 7.10 PDF

• Ex. 7.1


• Ex. 7.2


• Ex. 7.2


• Ex. 7.3


• Ex. 7.4


• Ex. 7.5


• Ex. 7.6


• Ex. 7.7


• Ex. 7.8


• Ex. 7.9


• Ex. 7.10


• Ex. 7.11


• Miscellaneous Exercise

Question 1. Evaluate the integral: \( \int_{0}^{1} \frac{x}{x^2 + 1} \, dx \)

Step 1: Substitution Let \( x^2 + 1 = t \).
Differentiating both sides: \( 2x \, dx = dt \implies x \, dx = \frac{dt}{2} \).

Step 2: Change Limits When \( x = 0 \), \( t = 0^2 + 1 = 1 \).
When \( x = 1 \), \( t = 1^2 + 1 = 2 \).

Step 3: Integrate Substituting values into the integral: \[ \begin{align} I &= \int_{1}^{2} \frac{1}{t} \cdot \frac{dt}{2} \\ &= \frac{1}{2} \int_{1}^{2} \frac{dt}{t} \\ &= \frac{1}{2} [\log |t|]_{1}^{2} \\ &= \frac{1}{2} (\log 2 - \log 1) \end{align} \] Since \( \log 1 = 0 \): \[ I = \frac{1}{2} \log 2 \]

Question 2. Evaluate the integral: \( \int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos^5 \phi \, d\phi \)

Step 1: Simplify the Integrand Rewrite \( \cos^5 \phi \) as \( \cos \phi \cdot (\cos^2 \phi)^2 \): \[ I = \int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} (1 - \sin^2 \phi)^2 \cos \phi \, d\phi \]

Step 2: Substitution Let \( \sin \phi = t \).
Then \( \cos \phi \, d\phi = dt \).
Limits: When \( \phi = 0, t = 0 \). When \( \phi = \frac{\pi}{2}, t = 1 \).

Step 3: Integrate \[ \begin{align} I &= \int_{0}^{1} \sqrt{t} (1 - t^2)^2 \, dt \\ &= \int_{0}^{1} t^{1/2} (1 - 2t^2 + t^4) \, dt \\ &= \int_{0}^{1} (t^{1/2} - 2t^{5/2} + t^{9/2}) \, dt \\ &= \left[ \frac{2}{3}t^{3/2} - 2\left(\frac{2}{7}\right)t^{7/2} + \frac{2}{11}t^{11/2} \right]_{0}^{1} \\ &= \left( \frac{2}{3} - \frac{4}{7} + \frac{2}{11} \right) - 0 \\ &= \frac{154 - 132 + 42}{231} \\ &= \frac{64}{231} \end{align} \]

Question 3. Evaluate the integral: \( \int_{0}^{1} \sin^{-1} \left( \frac{2x}{1+x^2} \right) \, dx \)

Step 1: Substitution Let \( x = \tan \theta \implies dx = \sec^2 \theta \, d\theta \).
Limits: When \( x=0, \theta=0 \). When \( x=1, \theta=\frac{\pi}{4} \).

Step 2: Simplify Integrand \( \frac{2x}{1+x^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta \).
So, \( \sin^{-1}(\sin 2\theta) = 2\theta \).

Step 3: Integration by Parts \[ I = \int_{0}^{\frac{\pi}{4}} 2\theta \sec^2 \theta \, d\theta \] Using integration by parts: \( \int u v' = uv - \int u'v \).
Take \( u = \theta \) and \( v' = \sec^2 \theta \). Then \( u' = 1 \) and \( v = \tan \theta \). \[ \begin{align} I &= 2 \left( [\theta \tan \theta]_{0}^{\frac{\pi}{4}} - \int_{0}^{\frac{\pi}{4}} \tan \theta \, d\theta \right) \\ &= 2 \left( \left(\frac{\pi}{4} \cdot 1 - 0\right) - [\log |\sec \theta|]_{0}^{\frac{\pi}{4}} \right) \\ &= 2 \left( \frac{\pi}{4} - (\log \sqrt{2} - \log 1) \right) \\ &= \frac{\pi}{2} - 2 \log (2^{1/2}) \\ &= \frac{\pi}{2} - \log 2 \end{align} \]

Question 4. Evaluate the integral: \( \int_{0}^{2} x\sqrt{x+2} \, dx \)

Step 1: Substitution Let \( x+2 = t^2 \implies dx = 2t \, dt \). Also \( x = t^2 - 2 \).
Limits: When \( x=0, t=\sqrt{2} \). When \( x=2, t=2 \).

Step 2: Integrate \[ \begin{align} I &= \int_{\sqrt{2}}^{2} (t^2 - 2) \cdot t \cdot 2t \, dt \\ &= 2 \int_{\sqrt{2}}^{2} (t^4 - 2t^2) \, dt \\ &= 2 \left[ \frac{t^5}{5} - \frac{2t^3}{3} \right]_{\sqrt{2}}^{2} \end{align} \]

Step 3: Evaluate Limits Upper limit (\(t=2\)): \( 2(\frac{32}{5} - \frac{16}{3}) = 2(\frac{96-80}{15}) = \frac{32}{15} \).
Lower limit (\(t=\sqrt{2}\)): \( 2(\frac{4\sqrt{2}}{5} - \frac{4\sqrt{2}}{3}) = 8\sqrt{2}(\frac{1}{5}-\frac{1}{3}) = 8\sqrt{2}(\frac{-2}{15}) = -\frac{16\sqrt{2}}{15} \).
Result: \[ I = \frac{32}{15} - \left( -\frac{16\sqrt{2}}{15} \right) = \frac{16\sqrt{2}(\sqrt{2}+1)}{15} \]

Question 5. Evaluate the integral: \( \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos^2 x} \, dx \)

Step 1: Substitution Let \( \cos x = t \implies -\sin x \, dx = dt \implies \sin x \, dx = -dt \).
Limits: \( x=0 \to t=1 \); \( x=\frac{\pi}{2} \to t=0 \).

Step 2: Integrate \[ \begin{align} I &= \int_{1}^{0} \frac{-dt}{1+t^2} \\ &= \int_{0}^{1} \frac{dt}{1+t^2} \\ &= [\tan^{-1} t]_{0}^{1} \\ &= \tan^{-1}(1) - \tan^{-1}(0) \\ &= \frac{\pi}{4} \end{align} \]

Question 6. Evaluate the integral: \( \int_{0}^{2} \frac{dx}{x+4-x^2} \)

Step 1: Complete the Square Denominator: \( -(x^2 - x - 4) = -[(x - \frac{1}{2})^2 - \frac{1}{4} - 4] = -[(x-\frac{1}{2})^2 - \frac{17}{4}] \)
\( = \left( \frac{\sqrt{17}}{2} \right)^2 - \left( x - \frac{1}{2} \right)^2 \).

Step 2: Apply Formula Use \( \int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| \).
Here \( a = \frac{\sqrt{17}}{2} \). \[ \begin{align} I &= \frac{1}{2(\frac{\sqrt{17}}{2})} \left[ \log \left| \frac{\frac{\sqrt{17}}{2} + (x-\frac{1}{2})}{\frac{\sqrt{17}}{2} - (x-\frac{1}{2})} \right| \right]_{0}^{2} \\ &= \frac{1}{\sqrt{17}} \left[ \log \left| \frac{\sqrt{17} + 2x - 1}{\sqrt{17} - 2x + 1} \right| \right]_{0}^{2} \end{align} \]

Step 3: Evaluate Limits Upper limit (\(x=2\)): \( \log \left| \frac{\sqrt{17} + 3}{\sqrt{17} - 3} \right| \).
Lower limit (\(x=0\)): \( \log \left| \frac{\sqrt{17} - 1}{\sqrt{17} + 1} \right| \).
Combining logarithms leads to: \[ I = \frac{1}{\sqrt{17}} \log \left( \frac{21 + 5\sqrt{17}}{4} \right) \]

Question 7. Evaluate the integral: \( \int_{-1}^{1} \frac{dx}{x^2+2x+5} \)

Step 1: Complete the Square \( x^2 + 2x + 5 = (x^2 + 2x + 1) + 4 = (x+1)^2 + 2^2 \).

Step 2: Integrate Using \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) \): \[ \begin{align} I &= \int_{-1}^{1} \frac{dx}{(x+1)^2 + 2^2} \\ &= \left[ \frac{1}{2} \tan^{-1}\left(\frac{x+1}{2}\right) \right]_{-1}^{1} \end{align} \]

Step 3: Evaluate \[ \begin{align} I &= \frac{1}{2} \left( \tan^{-1}(1) - \tan^{-1}(0) \right) \\ &= \frac{1}{2} \left( \frac{\pi}{4} - 0 \right) \\ &= \frac{\pi}{8} \end{align} \]

Question 8. Evaluate the integral: \( \int_{1}^{2} \left( \frac{1}{x} - \frac{1}{2x^2} \right) e^{2x} \, dx \)

Step 1: Substitution Let \( 2x = t \implies x = t/2 \implies dx = dt/2 \).
Limits: \( x=1 \to t=2 \); \( x=2 \to t=4 \).

Step 2: Transform Integral \[ \begin{align} I &= \int_{2}^{4} \left( \frac{1}{t/2} - \frac{1}{2(t/2)^2} \right) e^t \frac{dt}{2} \\ &= \int_{2}^{4} \left( \frac{2}{t} - \frac{2}{t^2} \right) e^t \frac{dt}{2} \\ &= \int_{2}^{4} e^t \left( \frac{1}{t} - \frac{1}{t^2} \right) \, dt \end{align} \]

Step 3: Apply Property We know \( \int e^x [f(x) + f'(x)] dx = e^x f(x) \).
Here \( f(t) = 1/t \) and \( f'(t) = -1/t^2 \). \[ \begin{align} I &= \left[ \frac{e^t}{t} \right]_{2}^{4} \\ &= \frac{e^4}{4} - \frac{e^2}{2} \\ &= \frac{e^2(e^2 - 2)}{4} \end{align} \]

Question 9. The value of the integral \( \int_{\frac{1}{3}}^{1} \frac{(x-x^3)^{\frac{1}{3}}}{x^4} \, dx \) is:

Step 1: Simplify Numerator Factor \( x^3 \) out of the bracket: \[ (x-x^3)^{1/3} = [x^3(x^{-2}-1)]^{1/3} = x(x^{-2}-1)^{1/3} \]

Step 2: Rewrite Integral \[ I = \int_{\frac{1}{3}}^{1} \frac{x(x^{-2}-1)^{1/3}}{x^4} \, dx = \int_{\frac{1}{3}}^{1} (x^{-2}-1)^{1/3} \cdot x^{-3} \, dx \]

Step 3: Substitution Let \( x^{-2} - 1 = t \).
Diff: \( -2x^{-3} \, dx = dt \implies x^{-3} \, dx = -\frac{1}{2} dt \).
Limits: \( x=\frac{1}{3} \implies t = 9-1=8 \); \( x=1 \implies t = 1-1=0 \).

Step 4: Evaluate \[ \begin{align} I &= \int_{8}^{0} t^{1/3} \left( -\frac{1}{2} \right) dt \\ &= \frac{1}{2} \int_{0}^{8} t^{1/3} \, dt \\ &= \frac{1}{2} \left[ \frac{3}{4} t^{4/3} \right]_{0}^{8} \\ &= \frac{3}{8} [16 - 0] \\ &= 6 \end{align} \]

Correct Option: A

Question 10. If \( f(x) = \int_{0}^{x} t \sin t \, dt \), then \( f'(x) \) is:

Step 1: Applying Fundamental Theorem of Calculus (Leibniz Rule) The derivative of an integral function \( F(x) = \int_{a}^{x} g(t) \, dt \) is simply \( g(x) \). \[ \frac{d}{dx} \left( \int_{0}^{x} t \sin t \, dt \right) = x \sin x \]

Alternatively, integrating by parts first: \[ f(x) = [-t \cos t + \sin t]_0^x = -x \cos x + \sin x \] Differentiating this: \[ f'(x) = -(\cos x - x \sin x) + \cos x = x \sin x \]

Correct Option: B

Class 12 Physics Formula Sheet Complete Chapterwise List

Complete Physics Formula Sheet - Class 12

A comprehensive pedagogical guide for quick revision.

Electric Charges and Fields

Properties of Charge

(i) Quantisation: \( q = ne \) (where \( n=0,1,2... \), \( e = 1.6 \times 10^{-19} C \))
(ii) Additivity: \( q_{net} = \sum q \)
(iii) Conservation: Total charge of an isolated system is constant.

Coulomb's Law

\( F = K \frac{q_1 q_2}{r^2} \)

Where \( K = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 \)

Vector form: \( \vec{F}_{12} = K \frac{q_1 q_2}{r_{12}^2} \hat{r}_{12} \) where \( \hat{r}_{12} = \frac{\vec{r}_{12}}{r_{12}} \)

Principle of Superposition

\( \vec{F}_1 = \vec{F}_{12} + \vec{F}_{13} + ... + \vec{F}_{1n} \)
\( \epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2 \)

Electric Field

\( \vec{E} = \frac{\vec{F}}{q_0} \) or \( \vec{E} = K \frac{q}{r^2} \)
Vector form: \( \vec{E} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r} \)

Relative Permittivity

\( \epsilon_r = \frac{\epsilon}{\epsilon_0} \) (Permittivity in medium / Permittivity in free space)

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• Physics - July 2017 View Answer Key

Charge Distributions & Gauss's Law

Charge Densities

(i) Linear: \( \lambda = dq/dl \)
(ii) Surface: \( \sigma = dq/dA \)
(iii) Volume: \( \rho = dq/dV \)

Electric Dipole

Dipole Moment: \( \vec{p} = q(2\vec{a}) \) (direction \( -q \) to \( +q \))
(i) At axial position: \( \vec{E}_{axial} = \frac{1}{4\pi\epsilon_0} \frac{2\vec{p}}{r^3} \)
(ii) At equatorial position: \( \vec{E}_{equa} = \frac{-1}{4\pi\epsilon_0} \frac{\vec{p}}{r^3} \)

Torque & Flux

Torque: \( \vec{\tau} = \vec{p} \times \vec{E} \) or \( \tau = pE \sin\theta \)
Electric Flux: \( \phi_E = \oint \vec{E} \cdot d\vec{A} = \oint E dA \cos\theta \)

Gauss Theorem & Applications

\( \phi_E = \frac{q}{\epsilon_0} = \oint \vec{E} \cdot d\vec{A} \)
(i) Infinitely long wire: \( E = \frac{\lambda}{2\pi\epsilon_0 r} \)
(ii) Infinite plane sheet: \( E = \frac{\sigma}{2\epsilon_0} \)

Applications of Gauss's Law (Spheres)

(iii) Uniformly charged thin spherical shell

(a) Outside (\(r > R\)): \( E_{out} = \frac{\sigma R^2}{\epsilon_0 r^2} \)
(b) At surface (\(r = R\)): \( E_{surf} = \frac{\sigma}{\epsilon_0} \)
(c) Internal point (\(r < R\)): \( E_{in} = 0 \)

(iv) Solid non-conducting sphere

(a) Outside (\(r > R\)): \( E_{out} = \frac{\rho R^3}{3\epsilon_0 r^2} \)
(b) At surface (\(r = R\)): \( E_{surf} = \frac{\rho R}{3\epsilon_0} \)
(c) Internal Point (\(r < R\)): \( E_{in} = \frac{\rho r}{3\epsilon_0} \)

Electrostatic Potential & Capacitance

Potential

\( V = \frac{W}{q_0} \); Work \( W = q \times \Delta V \)
Relation to Field: \( V = - \int_{\infty}^r \vec{E} \cdot d\vec{r} \)
Due to point charge: \( V = \frac{1}{4\pi\epsilon_0} \frac{q}{r} \)

Potential due to Dipole

(a) End on: \( V = \frac{1}{4\pi\epsilon_0} \frac{p}{r^2} \)
(b) Equatorial: \( V = 0 \)
(c) Any point: \( V = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2} \)

Work & Energy (Dipole)

Work to rotate: \( W = pE(\cos\theta_0 - \cos\theta) \)
Potential Energy: \( U_{\theta} = -pE \cos\theta = -\vec{p} \cdot \vec{E} \)

Capacitance

\( C = \frac{q}{V} \)
Isolated Spherical Conductor: \( C = 4\pi\epsilon_0 K a \)
Parallel Plate Capacitor: \( C = \frac{K \epsilon_0 A}{d} \) (For air \( K=1 \), \( C_0 = \frac{\epsilon_0 A}{d} \))
Force between plates: \( F = \frac{1}{2} qE \)

Energy of Charged Conductor

\( U = \frac{1}{2} qV \) or \( U = \frac{1}{2} \frac{q^2}{C} \) or \( U = \frac{1}{2} CV^2 \)

Capacitance (Continued)

Formulas

Energy Density: \( u = \frac{U}{Ad} = \frac{1}{2} \epsilon_0 E^2 \)
With Dielectric Slab: \( C = \frac{\epsilon_0 A}{(d-t) + \frac{t}{K}} \)
Series Combination: \( \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \)
Parallel Combination: \( C = C_1 + C_2 + C_3 \)

Current Electricity

Current: \( I = \frac{q}{t} \) or \( I(t) = \lim_{\Delta t \to 0} \frac{\Delta Q}{\Delta t} \)
Ohm's Law: \( V = RI \)
Resistivity: \( \rho = \frac{RA}{l} \) or \( \rho = \frac{m}{ne^2\tau} \)
Drift Velocity: \( v_d = \frac{eE\tau}{m} \) or \( v_d = \left( \frac{eV}{ml} \right) \tau \)
Current Density: \( j = i/A \) or \( \vec{j} = n e \vec{v}_d \)
Mobility: \( \mu = \frac{v_d}{E} \)
Vector form of Ohm's Law: \( \vec{j} = \sigma \vec{E} \)

Resistance & Temperature

\( R_t = R_0(1 + \alpha t) \) where \(\alpha\) is temp coefficient.

Combination

(A) Series: \( R = R_1 + R_2 + R_3 \)
(B) Parallel: \( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \)

Power & Energy

Power \( P = i^2 R = V^2/R \)
Heat \( H = i^2 R t \)

Current Electricity (Circuits)

Terminal Voltage: \( V = E - ir \)
Internal Resistance: \( r = R \left[ \frac{E}{V} - 1 \right] \)
Kirchhoff's Laws: 1. \( \sum i = 0 \) (Junction rule)
2. \( \sum iR = \sum E \) (Loop rule)
Wheatstone Bridge: \( \frac{P}{Q} = \frac{R}{S} \)
Meter Bridge: \( S = R \left[ \frac{100-l}{l} \right] \)
Potentiometer: \( r = R \left( \frac{l_1}{l_2} - 1 \right) \)

Moving Charges and Magnetism

Laws & Fields

Biot Savart Law: \( d\vec{B} = \frac{\mu_0}{4\pi} \frac{i (d\vec{l} \times \vec{r})}{r^3} \)
Relation: \( c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \)
Circular Coil (Axis): \( B = \frac{\mu_0 N I a^2}{2(a^2 + x^2)^{3/2}} \)
Ampere's Circuital Law: \( \oint \vec{B} \cdot d\vec{l} = \mu_0 i \)
Infinite Straight Wire: \( B = \frac{\mu_0 i}{2\pi r} \)
Solenoid: \( B = \mu_0 n i \); Toroid: \( B = \mu_0 n i \)

Lorentz Force

\( \vec{F} = q(\vec{v} \times \vec{B}) \)
Cyclotron: \( r = \frac{mv}{qB} \), \( T = \frac{2\pi m}{qB} \), \( K_{max} = \frac{q^2 B^2 R^2}{2m} \)

Forces & Instruments

Force on Conductor: \( F = i B L \sin\theta \)
Field at center of loop: \( B = \frac{\mu_0 i}{2a} \) (or \( \frac{\mu_0 N i}{2a} \))
Straight conductor (finite): \( B = \frac{\mu_0 i}{4\pi r}(\sin\phi_1 + \sin\phi_2) \)
Torque on Bar Magnet: \( \vec{\tau} = \vec{M} \times \vec{B} = MB \sin\theta \)
Potential Energy: \( U = -MB \cos\theta \)
Moving Coil Galvanometer: Deflection \( \phi = \left( \frac{NAB}{K} \right) I \)
Bohr Magneton: \( M_{min} = \frac{eh}{4\pi m_e} \)

Magnetism and Matter

Bar Magnet (Solenoid equiv): \( B = \frac{\mu_0 2M}{4\pi r^3} \)
Earth's Field: \( B_E = \sqrt{B_H^2 + B_V^2} \), \( \theta = \tan^{-1} \frac{B_V}{B_H} \)
Magnetisation: \( \vec{I} = \frac{\vec{M}}{V} \)
Magnetic Intensity: \( \vec{H} = \frac{\vec{B}}{\mu_0} - \vec{I} \)
Relative Permeability: \( \mu_r = 1 + \chi_m \)
Curie's Law: \( I = C \left( \frac{H}{T} \right) \)
Gauss Law for Magnetism: \( \oint \vec{B} \cdot d\vec{A} = 0 \)

Electromagnetic Induction

Magnetic Flux: \( \phi_B = \vec{B} \cdot \vec{A} = BA \cos\theta \)
Induced EMF: \( e = - \frac{d\phi_B}{dt} \) (for N turns \( e = -N \frac{d\phi}{dt} \))
Induced Current: \( i = \frac{e}{R} \)
Motional EMF: \( e = Bvl \)
Self Inductance: \( L = \frac{N \phi_B}{i} \); Solenoid \( L = \frac{\mu_0 N^2 A}{l} \)
Energy stored: \( U = \frac{1}{2} L i_0^2 \)
Mutual Inductance: \( M = \frac{\mu_0 N_1 N_2 A}{l} \)

Alternating Current

Mean Value: \( i_m = \frac{2}{\pi} i_0 = 0.637 i_0 \)
RMS Value: \( i_{rms} = \frac{i_0}{\sqrt{2}} = 0.707 i_0 \)
Reactance: \( X_L = \omega L \), \( X_C = \frac{1}{\omega C} \)
Impedance (LCR): \( Z = \sqrt{R^2 + (X_L - X_C)^2} \)
Resonance Frequency: \( f = \frac{1}{2\pi \sqrt{LC}} \)
Power Factor: \( \cos\phi = \frac{R}{Z} \)
Transformer: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} = r \)

Electromagnetic Waves

Displacement Current: \( i_d = \epsilon_0 \frac{d\phi_E}{dt} \)
Maxwell's Equations: 1. \( \oint \vec{E} \cdot d\vec{A} = q/\epsilon_0 \)
2. \( \oint \vec{B} \cdot d\vec{A} = 0 \)
3. \( \oint \vec{E} \cdot d\vec{l} = -d\phi_B/dt \)
4. \( \oint \vec{B} \cdot d\vec{l} = \mu_0(i + i_d) \)
Relation: \( c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \), \( \frac{E}{B} = c \)

Ray Optics

Mirror Equation: \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \), \( f = R/2 \)
Snell's Law: \( \frac{\sin i}{\sin r} = {}_1n_2 = \frac{n_2}{n_1} \)
Lens Formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
Power: \( P = 1/f \)
Prism: \( n = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)} \)
Refraction at Spherical Surface: \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \)

Wave Optics

Interference: \( I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos\phi \)
Bright Fringes: \( x = m \frac{D\lambda}{d} \)
Fringe Width: \( W = \frac{D\lambda}{d} \)
Diffraction (Single slit): \( e \sin\theta = \pm m\lambda \) (minima)
Brewster's Law: \( \mu = \tan i_B \)

Dual Nature of Radiation

Einstein's Photoelectric Eq: \( K_{max} = h\nu - \phi_0 = h(\nu - \nu_0) \)
Energy: \( E = h\nu = hc/\lambda \)
De Broglie Wavelength: \( \lambda = \frac{h}{p} = \frac{h}{mv} = \frac{h}{\sqrt{2mK}} \)
Heisenberg's Uncertainty: \( \Delta x \cdot \Delta p \approx \hbar \)

Atoms

Bohr Quantization: \( mvr = \frac{nh}{2\pi} \)
Radius: \( r_n \propto n^2 \)
Energy: \( E_n = - \frac{13.6}{n^2} \, \text{eV} \)
Rydberg Formula: \( \frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] \)

Nuclei

Size: \( R = R_0 A^{1/3} \)
Mass Energy: \( E = mc^2 \)
Radioactive Decay: \( N = N_0 e^{-\lambda t} \)
Half Life: \( T_{1/2} = \frac{0.6931}{\lambda} \)

Semiconductors

\( n_e n_h = n_i^2 \)
Gains: \( \alpha = \frac{i_c}{i_e} \), \( \beta = \frac{i_c}{i_b} \)
Relation: \( \alpha = \frac{\beta}{1+\beta} \)

Logic Gates (Summary)

Gate	Boolean Expression
OR	\( Y = A + B \)
AND	\( Y = A \cdot B \)
NOT	\( Y = \bar{A} \)
NAND	\( Y = \overline{A \cdot B} \)
NOR	\( Y = \overline{A + B} \)

Class X Social Science Important Map works.

Class 10 Social Science Map Work

Complete Syllabus & Revision PDF (2024-25)

Download Map Work PDF ↓

This PDF covers the complete map work syllabus for Class 10 History and Geography. It is designed to help students easily identify, locate, and label key places on the map of India for the board exams.

HISTORY: Nationalism in India

1. Indian National Congress Sessions

• Calcutta (September 1920)
• Nagpur (December 1920)
• Madras (1927)

2. Important Centres of National Movement

• Champaran (Bihar) - Indigo Planters Movement
• Kheda (Gujarat) - Peasant Satyagraha
• Ahmedabad (Gujarat) - Cotton Mill Workers Satyagraha
• Amritsar (Punjab) - Jallianwala Bagh Incident
• Chauri Chaura (U.P.) - Calling off Non-Cooperation Movement
• Dandi (Gujarat) - Civil Disobedience Movement

GEOGRAPHY: Outline Political Map

1. Resources and Development

• Identification of Major Soil Types found in India.

2. Water Resources (Dams)

Locating and Labelling:

• Salal, Bhakra Nangal, Tehri
• Rana Pratap Sagar, Sardar Sarovar
• Hirakud, Nagarjuna Sagar, Tungabhadra

3. Agriculture

• Major Areas: Rice and Wheat.
• Largest Producer States: Sugarcane, Tea, Coffee, Rubber, Cotton, and Jute.

4. Minerals and Energy Resources

• Iron Ore Mines: Mayurbhanj, Durg, Bailadila, Bellary, Kudremukh.
• Coal Mines: Raniganj, Bokaro, Talcher, Neyveli.
• Oil Fields: Digboi, Naharkatia, Mumbai High, Bassien, Kalol, Ankaleshwar.
• Thermal Power Plants: Namrup, Singrauli, Ramagundam.
• Nuclear Power Plants: Narora, Kakrapara, Tarapur, Kalpakkam.

5. Manufacturing Industries

• Cotton Textile: Mumbai, Indore, Surat, Kanpur, Coimbatore.
• Iron and Steel: Durgapur, Bokaro, Jamshedpur, Bhilai, Vijayanagar, Salem.
• Software Technology Parks: Noida, Gandhinagar, Mumbai, Pune, Hyderabad, Bengaluru, Chennai, Thiruvananthapuram.

6. Lifelines of National Economy

• Major Ports: Kandla, Mumbai, Marmagao, New Mangalore, Kochi, Tuticorin, Chennai, Visakhapatnam, Paradip, Haldia.
• International Airports:
  • Amritsar (Raja Sansi)
  • Delhi (Indira Gandhi)
  • Mumbai (Chhatrapati Shivaji)
  • Chennai (Meenam Bakkam)
  • Kolkata (Netaji Subhash Chandra Bose)
  • Hyderabad (Rajiv Gandhi)

Mastering 3D Shapes: A Guide for 3rd Grade Maths | Geometry Term 1 Unit 1

Construction of 3D shapes

3rd Maths : Term 1 Unit 1 : Geometry

Teacher’s note : Cuboid, Cube, Cylinder, Sphere, Pyramid, Cone - Teacher can guide the children to draw 3D object and give the real object of the 3D shapes, so that the children can touch, feel and find out the answers.

Construction of 3D shapes.

Look at the things around you.

Identify the shapes of the objects. Observe the dimensions of these objects.

Cuboid, Cube, Cylinder, Sphere, Pyramid, Cone

3D Shapes

Cuboid, Cube, Cylinder, Sphere, Pyramid, Cone

Solid shapes have 3 Dimensions namely length, breadth, and height.

These are shortly called as 3D shapes.

Let us know

A Cube is a solid shape made of squares. It has 6 faces, 12 edges and 8 vertices.

A Cuboid is a solid shape made of rectangles. It has 6 faces, 12 edges and 8 vertices.

A Sphere is a solid shape made of circles. It has 1 faces, no edges and no vertex.

Practice

1. Complete the following table by filing the properties of 2d and 3d shapes.

2. Match the following.

Click to see the answers

Sphere → Ball

Cylinder → Can

Cone → Birthday Cap

Cube → Dice

Cuboid → Book

Teacher’s note : Teacher shall facilitate the children to draw the front and side views of 3d shapes by providing the objects.

Activity 4

Join the dots to form 3D shapes.

Activity 5

Complete solid shapes and colour it.

Sphere, Cuboid, Cube, Cone, Cylinder, Cube

Teacher’s note : Teacher can guide the children to draw 3D object and give the real object of the 3D shapes, so that the children can touch, feel and find out the answers.

Mastering 2D Shapes: A Guide for 3rd Grade Maths (Term 1, Geometry)

Construction of 2D shapes

3rd Maths | Term 1, Unit 1: Geometry

Understanding Properties of 2D Shapes

Let us understand the properties of 2D shapes.

Square

Let us know

A Square has four sides. All the four sides are equal.

Let us make a square by folding a paper by following the given steps.

Step 1: Take a paper

Step 2: Fold the paper as shown in the figure.

Shade the extra portion in the bottom with red colour. Red coloured portion will be rectangular in shape. Tear it off and keep it aside. Open up the triangle. What do you observe? You could see a square.

The crease in the middle of the square is called the 'Diagonal of the square'.

You can note that the diagonal divides the square into two triangles.

Try this

Can you find the other diagonal of the square by folding it the other way? If so how many diagonals can you find for a square?.

Teacher's note:

Teacher can guide the children to do this paper folding activity.

Observe the number of sides and corners of a square.

So, a square has four sides, four corners and two diagonals.

Think

Are all the sides of a square equal

What about the diagonals?

Are they equal?

We know from square shape

we shall summarize the properties of a square as follows

(i) Square has four sides.

(ii) All the four sides are equal.

(iii) Square has four corners.

(iv) Square has two diagonals.

(v) The two diagonals are equal.

Rectangle

Step 1: Take the rectangular piece which was kept aside. Observe its sides.

Let us know:

Opposite sides are equal

Fold the opposite sides of the rectangle. What do you observe? The sides coincide.

Now we get opposite sides equal. Hence in a rectangle, opposite sides are equal.

Fold the opposite corners as we did in the square. Observe the crease. It shows the diagonal of the rectangle.

Let us know:

Diagonals are equal in rectangle.

We know from rectangle shape

The properties of a rectangle are as follows

(i) Rectangle has four sides.

(ii) Two opposite sides are equal.

(iii) Rectangle has four corners.

(iv) Rectangle has two diagonals.

(v) Two diagonals are equal.

Triangle

Fold the square along any of these diagonals to form a triangle.

Observe the Sides and corners of the triangle.

A triangle has three sides and three corners.

Cut the paper and make triangles of different kind.

Observe the length of the sides of the triangle. length of the sides. Let the children explore the names of different sides of the triangles.

Try This:

How many triangles can be made out of this square paper?

Fold the square along any of diagonals to form a triangle.

Let us know

Teacher’s note:

Facilitate the children to explore the properties of shapes in various aspects.

Circle

Let us know:

Circle is a closed curve

Draw a circle using pencil and Bangle.

Step 1: Place a bangle on the paper as shown in figure.

Step 2: Trace the outline of the bangle with a curved line with the pencil until you reach the starting point, we get a circle.

Now, we get a circle

On observing the circle drawn we shall write the properties of it as follows.

(i) Circle has no sides.

(ii) Circle has no corners.

(iii) Circle has a center point.

Activity 1

Write the names of few objects in everyday use and mention their geometrical shapes. Example, table− cuboid

cookies – Circle - Cylinder

wheels of a bike − Circle – Cylinder

ball – sphere

pencil - cylinder

clock faces − Circle

dinner plates − Circle

chess boards − Square

slices of pizza − Triangle

chapter book covers − Rectangle

cell phones – Rectangle - cuboid

Practice

1) Triangle has Three corners.

2) Four sides of a square are Equal.

3) Circle has No sides.

4) Rectangle has Two diagonals.

5) Opposite sides of a rectangle are Equal.

6) Circle has One centre point.

Plane Surface

Surface of few objects like walls, floors papers and top of a table are flat. Flat surfaces are otherwise called as plane surfaces or planes. Cubes and cuboids have flat surfaces.

Curved Surface

Surfaces of few objects such as ball, flowerwase, pot are curved. Cone, Cylinder and sphere have curved surfaces.

Activity 2

Draw all the 2d shapes in the given dot grid. One is done for you.

A geoboard is a mathematical manipulative board.

Join the dots in the grid using curved lines to make designs of your choice. one is done for you.

Teacher’s note :

Teacher can lead the children to make the shapes drawn by them in the dot grid by using rubber band in the Geo board drawn by them in the dot grid

Activity 3

We can see many things around us have straight lines and curved lines.

Draw any 5 shapes and put a tick in the given boxes to indicate the type of the lines found in them.

Teacher’s note :

Teacher can discuss about the types of lines found in objects in everyday use and enable the children to draw them in above tabular column.

Practice

Put a tick mark in the appropriate columns.