Integrals Class 12 Exercise 7.11 Solutions | Maths Part II CBSE

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Integrals Class 12 Mathematics Part II CBSE Solution - Exercise 7.11

Detailed solutions for Definite Integrals using properties (Exercise 7.11).

Question 1. By using the properties of definite integrals, evaluate the integral: \(\int_{0}^{\pi/2} \cos^2 x \, dx\)

Let \( I = \int_{0}^{\pi/2} \cos^2 x \, dx \quad \dots(1) \)

Using the property \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx \):

\( I = \int_{0}^{\pi/2} \cos^2(\frac{\pi}{2} - x) \, dx \)
\( \Rightarrow I = \int_{0}^{\pi/2} \sin^2 x \, dx \quad \dots(2) \)

Adding (1) and (2):

\( 2I = \int_{0}^{\pi/2} (\cos^2 x + \sin^2 x) \, dx \)
\( 2I = \int_{0}^{\pi/2} 1 \, dx \)
\( 2I = [x]_{0}^{\pi/2} = \frac{\pi}{2} - 0 \)
\( I = \frac{\pi}{4} \)

Question 2. Evaluate: \(\int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx\)

Let \( I = \int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \quad \dots(1) \)

Using the property \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx \):

\( I = \int_{0}^{\pi/2} \frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)} + \sqrt{\cos(\frac{\pi}{2}-x)}} \, dx \)
\( I = \int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx \quad \dots(2) \)

Adding (1) and (2):

\( 2I = \int_{0}^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \)
\( 2I = \int_{0}^{\pi/2} 1 \, dx \)
\( 2I = [x]_{0}^{\pi/2} = \frac{\pi}{2} \)
\( I = \frac{\pi}{4} \)

Question 3. Evaluate: \(\int_{0}^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} \, dx\)

Let \( I = \int_{0}^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} \, dx \quad \dots(1) \)

Using the property \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx \):

\( I = \int_{0}^{\pi/2} \frac{\sin^{3/2}(\frac{\pi}{2}-x)}{\sin^{3/2}(\frac{\pi}{2}-x) + \cos^{3/2}(\frac{\pi}{2}-x)} \, dx \)
\( I = \int_{0}^{\pi/2} \frac{\cos^{3/2} x}{\cos^{3/2} x + \sin^{3/2} x} \, dx \quad \dots(2) \)

Adding (1) and (2):

\( 2I = \int_{0}^{\pi/2} 1 \, dx \)
\( 2I = \frac{\pi}{2} \Rightarrow I = \frac{\pi}{4} \)

Question 4. Evaluate: \(\int_{0}^{\pi/2} \frac{\cos^5 x}{\sin^5 x + \cos^5 x} \, dx\)

Let \( I = \int_{0}^{\pi/2} \frac{\cos^5 x}{\sin^5 x + \cos^5 x} \, dx \quad \dots(1) \)

Using the property \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx \):

\( I = \int_{0}^{\pi/2} \frac{\cos^5(\frac{\pi}{2}-x)}{\sin^5(\frac{\pi}{2}-x) + \cos^5(\frac{\pi}{2}-x)} \, dx \)
\( I = \int_{0}^{\pi/2} \frac{\sin^5 x}{\cos^5 x + \sin^5 x} \, dx \quad \dots(2) \)

Adding (1) and (2):

\( 2I = \int_{0}^{\pi/2} 1 \, dx = \frac{\pi}{2} \)
\( I = \frac{\pi}{4} \)

Question 5. Evaluate: \(\int_{-5}^{5} |x+2| \, dx\)

We split the integral at \( x = -2 \) because \( |x+2| \) changes definition there.

\( |x+2| = -(x+2) \) for \( x < -2 \) and \( (x+2) \) for \( x \ge -2 \).

\( I = \int_{-5}^{-2} -(x+2) \, dx + \int_{-2}^{5} (x+2) \, dx \)
\( I = -[\frac{x^2}{2} + 2x]_{-5}^{-2} + [\frac{x^2}{2} + 2x]_{-2}^{5} \)

Calculating the first part:

\( -[(\frac{4}{2} - 4) - (\frac{25}{2} - 10)] = -[(2-4) - (12.5-10)] = -[-2 - 2.5] = 4.5 \)

Calculating the second part:

\( [(\frac{25}{2} + 10) - (\frac{4}{2} - 4)] = [(12.5+10) - (2-4)] = [22.5 - (-2)] = 24.5 \)

\( I = 4.5 + 24.5 = 29 \)

Question 6. Evaluate: \(\int_{2}^{8} |x-5| \, dx\)

Split the integral at \( x = 5 \).

\( |x-5| = -(x-5) \) for \( x < 5 \) and \( (x-5) \) for \( x \ge 5 \).

\( I = \int_{2}^{5} -(x-5) \, dx + \int_{5}^{8} (x-5) \, dx \)
\( I = -[\frac{x^2}{2} - 5x]_{2}^{5} + [\frac{x^2}{2} - 5x]_{5}^{8} \)

\( I = -[(\frac{25}{2}-25) - (\frac{4}{2}-10)] + [(\frac{64}{2}-40) - (\frac{25}{2}-25)] \)
\( I = -[-12.5 - (-8)] + [-8 - (-12.5)] \)
\( I = -[-4.5] + [4.5] = 4.5 + 4.5 = 9 \)

Answer: \( I = 9 \)

Question 7. Evaluate: \(\int_{0}^{1} x(1-x)^n \, dx\)

Using property \( \int_{0}^{a} f(x)dx = \int_{0}^{a} f(a-x)dx \):

\( I = \int_{0}^{1} (1-x)(1-(1-x))^n \, dx \)
\( I = \int_{0}^{1} (1-x)x^n \, dx \)
\( I = \int_{0}^{1} (x^n - x^{n+1}) \, dx \)

\( I = [\frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2}]_{0}^{1} \)
\( I = (\frac{1}{n+1} - \frac{1}{n+2}) - 0 \)
\( I = \frac{(n+2)-(n+1)}{(n+1)(n+2)} = \frac{1}{(n+1)(n+2)} \)

Question 8. Evaluate: \(\int_{0}^{\pi/4} \log(1+\tan x) \, dx\)

Let \( I = \int_{0}^{\pi/4} \log(1+\tan x) \, dx \quad \dots(1) \)

Using property \( x \to \frac{\pi}{4}-x \):

\( I = \int_{0}^{\pi/4} \log(1+\tan(\frac{\pi}{4}-x)) \, dx \)
Since \( \tan(A-B) = \frac{\tan A - \tan B}{1+\tan A \tan B} \), we have \( \tan(\frac{\pi}{4}-x) = \frac{1-\tan x}{1+\tan x} \).

\( I = \int_{0}^{\pi/4} \log(1 + \frac{1-\tan x}{1+\tan x}) \, dx \)
\( I = \int_{0}^{\pi/4} \log(\frac{1+\tan x + 1 - \tan x}{1+\tan x}) \, dx \)
\( I = \int_{0}^{\pi/4} \log(\frac{2}{1+\tan x}) \, dx \)
\( I = \int_{0}^{\pi/4} (\log 2 - \log(1+\tan x)) \, dx \)

\( I = \int_{0}^{\pi/4} \log 2 \, dx - I \)
\( 2I = \log 2 [x]_{0}^{\pi/4} = \frac{\pi}{4} \log 2 \)
\( I = \frac{\pi}{8} \log 2 \)

Question 9. Evaluate: \(\int_{0}^{2} x\sqrt{2-x} \, dx\)

Using property \( x \to 2-x \):

\( I = \int_{0}^{2} (2-x)\sqrt{2-(2-x)} \, dx \)
\( I = \int_{0}^{2} (2-x)\sqrt{x} \, dx \)
\( I = \int_{0}^{2} (2x^{1/2} - x^{3/2}) \, dx \)

\( I = [2\frac{x^{3/2}}{3/2} - \frac{x^{5/2}}{5/2}]_{0}^{2} \)
\( I = [\frac{4}{3}x^{3/2} - \frac{2}{5}x^{5/2}]_{0}^{2} \)
\( I = \frac{4}{3}(2\sqrt{2}) - \frac{2}{5}(4\sqrt{2}) \)
\( I = \frac{8\sqrt{2}}{3} - \frac{8\sqrt{2}}{5} = 8\sqrt{2}(\frac{5-3}{15}) = \frac{16\sqrt{2}}{15} \)

Question 10. Evaluate: \(\int_{0}^{\pi/2} (2\log \sin x - \log \sin 2x) \, dx\)

\( I = \int_{0}^{\pi/2} (2\log \sin x - \log(2\sin x \cos x)) \, dx \)
\( I = \int_{0}^{\pi/2} (2\log \sin x - \log 2 - \log \sin x - \log \cos x) \, dx \)
\( I = \int_{0}^{\pi/2} (\log \sin x - \log \cos x - \log 2) \, dx \quad \dots(1) \)

Using property \( P_4 \):

\( I = \int_{0}^{\pi/2} (\log \cos x - \log \sin x - \log 2) \, dx \quad \dots(2) \)

Adding (1) and (2):

\( 2I = \int_{0}^{\pi/2} (-2\log 2) \, dx \)
\( 2I = -2\log 2 [x]_{0}^{\pi/2} = -\pi \log 2 \)
\( I = -\frac{\pi}{2} \log 2 \) or \( \frac{\pi}{2} \log(\frac{1}{2}) \)

Question 11. Evaluate: \(\int_{-\pi/2}^{\pi/2} \sin^2 x \, dx\)

Let \( f(x) = \sin^2 x \). Since \( f(-x) = \sin^2(-x) = (-\sin x)^2 = \sin^2 x = f(x) \), \( f(x) \) is an even function.

Using the property for even functions:

\( I = 2 \int_{0}^{\pi/2} \sin^2 x \, dx \)

From Question 1, we know \( \int_{0}^{\pi/2} \sin^2 x \, dx = \frac{\pi}{4} \).

\( I = 2(\frac{\pi}{4}) = \frac{\pi}{2} \)

Question 12. Evaluate: \(\int_{0}^{\pi} \frac{x}{1+\sin x} \, dx\)

Using property \( x \to \pi - x \):

\( I = \int_{0}^{\pi} \frac{\pi - x}{1+\sin(\pi-x)} \, dx = \int_{0}^{\pi} \frac{\pi - x}{1+\sin x} \, dx \)

\( 2I = \int_{0}^{\pi} \frac{x + \pi - x}{1+\sin x} \, dx = \pi \int_{0}^{\pi} \frac{1}{1+\sin x} \, dx \)

Multiply numerator and denominator by \( 1-\sin x \):

\( 2I = \pi \int_{0}^{\pi} \frac{1-\sin x}{\cos^2 x} \, dx = \pi \int_{0}^{\pi} (\sec^2 x - \tan x \sec x) \, dx \)
\( 2I = \pi [\tan x - \sec x]_{0}^{\pi} \)
\( 2I = \pi [(\tan \pi - \sec \pi) - (\tan 0 - \sec 0)] \)
\( 2I = \pi [(0 - (-1)) - (0 - 1)] = \pi [1+1] = 2\pi \)
\( I = \pi \)

Question 13. Evaluate: \(\int_{-\pi/2}^{\pi/2} \sin^7 x \, dx\)

Let \( f(x) = \sin^7 x \).

\( f(-x) = \sin^7(-x) = (-\sin x)^7 = -\sin^7 x = -f(x) \).

Since \( f(x) \) is an odd function, the integral over \( [-a, a] \) is zero.

\( I = 0 \)

Question 14. Evaluate: \(\int_{0}^{2\pi} \cos^5 x \, dx\)

Let \( f(x) = \cos^5 x \).

\( f(2\pi - x) = \cos^5(2\pi - x) = \cos^5 x = f(x) \).
So, \( I = 2 \int_{0}^{\pi} \cos^5 x \, dx \).

Now consider the integral over \( [0, \pi] \). Let \( g(x) = \cos^5 x \).

\( g(\pi - x) = \cos^5(\pi - x) = (-\cos x)^5 = -\cos^5 x = -g(x) \).

Thus, \( \int_{0}^{\pi} \cos^5 x \, dx = 0 \).

Answer: \( I = 0 \)

Question 15. Evaluate: \(\int_{0}^{\pi/2} \frac{\sin x - \cos x}{1+\sin x \cos x} \, dx\)

Using property \( x \to \frac{\pi}{2} - x \):

\( I = \int_{0}^{\pi/2} \frac{\sin(\frac{\pi}{2}-x) - \cos(\frac{\pi}{2}-x)}{1 + \sin(\frac{\pi}{2}-x)\cos(\frac{\pi}{2}-x)} \, dx \)
\( I = \int_{0}^{\pi/2} \frac{\cos x - \sin x}{1 + \cos x \sin x} \, dx \)
\( I = - \int_{0}^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx \)

\( I = -I \Rightarrow 2I = 0 \Rightarrow I = 0 \)

Question 16. Evaluate: \(\int_{0}^{\pi} \log(1+\cos x) \, dx\)

\( I = \int_{0}^{\pi} \log(1+\cos(\pi-x)) \, dx = \int_{0}^{\pi} \log(1-\cos x) \, dx \)
\( 2I = \int_{0}^{\pi} \log((1+\cos x)(1-\cos x)) \, dx = \int_{0}^{\pi} \log(1-\cos^2 x) \, dx \)
\( 2I = \int_{0}^{\pi} \log(\sin^2 x) \, dx = 2\int_{0}^{\pi} \log \sin x \, dx \)
\( I = \int_{0}^{\pi} \log \sin x \, dx \)

Using property \( f(2a-x)=f(x) \) where \( 2a=\pi \), we have \( \sin(\pi-x)=\sin x \).

\( I = 2\int_{0}^{\pi/2} \log \sin x \, dx \)

We know \( \int_{0}^{\pi/2} \log \sin x \, dx = -\frac{\pi}{2} \log 2 \).

\( I = 2(-\frac{\pi}{2} \log 2) = -\pi \log 2 \)

Question 17. Evaluate: \(\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} \, dx\)

Let \( I = \int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} \, dx \quad \dots(1) \)

Using property \( x \to a-x \):

\( I = \int_{0}^{a} \frac{\sqrt{a-x}}{\sqrt{a-x} + \sqrt{x}} \, dx \quad \dots(2) \)

Adding (1) and (2):

\( 2I = \int_{0}^{a} 1 \, dx = [x]_{0}^{a} = a \)
\( I = \frac{a}{2} \)

Question 18. Evaluate: \(\int_{0}^{4} |x-1| \, dx\)

Split at \( x=1 \).

\( I = \int_{0}^{1} -(x-1) \, dx + \int_{1}^{4} (x-1) \, dx \)
\( I = [x - \frac{x^2}{2}]_{0}^{1} + [\frac{x^2}{2} - x]_{1}^{4} \)

\( I = (1 - 0.5) + [(8-4) - (0.5-1)] \)
\( I = 0.5 + [4 + 0.5] = 5 \)

Question 19. Show that \(\int_{0}^{a} f(x)g(x) \, dx = 2\int_{0}^{a} f(x) \, dx\), if \( f \) and \( g \) are defined as \( f(x)=f(a-x) \) and \( g(x)+g(a-x)=4 \).

Let \( I = \int_{0}^{a} f(x)g(x) \, dx \).

Using property \( x \to a-x \):

\( I = \int_{0}^{a} f(a-x)g(a-x) \, dx \)
Since \( f(a-x) = f(x) \) and \( g(a-x) = 4 - g(x) \):
\( I = \int_{0}^{a} f(x)(4 - g(x)) \, dx \)
\( I = 4\int_{0}^{a} f(x) \, dx - \int_{0}^{a} f(x)g(x) \, dx \)
\( I = 4\int_{0}^{a} f(x) \, dx - I \)

\( 2I = 4\int_{0}^{a} f(x) \, dx \)
\( I = 2\int_{0}^{a} f(x) \, dx \)

Hence proved.

Question 20. The value of \(\int_{-\pi/2}^{\pi/2} (x^3 + x\cos x + \tan^5 x + 1) \, dx\) is

Let \( f(x) = x^3 + x\cos x + \tan^5 x + 1 \).

\( \int_{-\pi/2}^{\pi/2} x^3 \, dx = 0 \) (Odd function)
\( \int_{-\pi/2}^{\pi/2} x\cos x \, dx = 0 \) (Odd function)
\( \int_{-\pi/2}^{\pi/2} \tan^5 x \, dx = 0 \) (Odd function)
\( \int_{-\pi/2}^{\pi/2} 1 \, dx = [x]_{-\pi/2}^{\pi/2} = \pi \)

Total value = \( \pi \).

C. \(\pi\)

Question 21. The value of \(\int_{0}^{\pi/2} \log(\frac{4+3\sin x}{4+3\cos x}) \, dx\) is

Let \( I = \int_{0}^{\pi/2} \log(\frac{4+3\sin x}{4+3\cos x}) \, dx \).

Using property \( x \to \frac{\pi}{2}-x \):

\( I = \int_{0}^{\pi/2} \log(\frac{4+3\sin(\pi/2-x)}{4+3\cos(\pi/2-x)}) \, dx \)
\( I = \int_{0}^{\pi/2} \log(\frac{4+3\cos x}{4+3\sin x}) \, dx \)
\( I = \int_{0}^{\pi/2} \log((\frac{4+3\sin x}{4+3\cos x})^{-1}) \, dx \)
\( I = -\int_{0}^{\pi/2} \log(\frac{4+3\sin x}{4+3\cos x}) \, dx \)
\( I = -I \)

\( 2I = 0 \Rightarrow I = 0 \)

C. 0