- 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, \[ \int \sin^2(2x+5) \, dx = \int \frac{1 - \cos 2(2x+5)}{2} \, dx \] \[ = \int \frac{1 - \cos(4x+10)}{2} \, dx \] \[ = \frac{1}{2} \int 1 \, dx - \frac{1}{2} \int \cos(4x+10) \, dx \] \[ = \frac{1}{2}x - \frac{1}{2} \cdot \frac{\sin(4x+10)}{4} + C \] \[ = \frac{x}{2} - \frac{1}{8}\sin(4x+10) + C \]
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 \(\cos^2 A = \frac{1+\cos 2A}{2}\): \[ = \frac{1}{2} \int \left[ \frac{1}{2}(\cos 12x + \cos 8x) + \frac{1+\cos 4x}{2} \right] \, dx \] \[ = \frac{1}{4} \int (\cos 12x + \cos 8x + 1 + \cos 4x) \, dx \] \[ = \frac{1}{4} \left[ \frac{\sin 12x}{12} + \frac{\sin 8x}{8} + x + \frac{\sin 4x}{4} \right] + C \]
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 \(t = \cos(2x+1)\): \[ = \frac{-1}{2} \cos(2x+1) + \frac{1}{6} \cos^3(2x+1) + C \]
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: \[ \int \frac{dt}{\cos^2 t} = \int \sec^2 t \, dt \] \[ = \tan t + C \] \[ = \tan(x e^x) + C \]
Differentiating using product rule: \((e^x \cdot 1 + x \cdot e^x) \, dx = dt\).
\(e^x(1+x) \, dx = dt\).
Substituting into integral: \[ \int \frac{dt}{\cos^2 t} = \int \sec^2 t \, dt \] \[ = \tan t + C \] \[ = \tan(x e^x) + C \]