Class 12-NCERT Solutions-Chapter-07-Integrals-Ex 7.10

Solution:

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

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

\[ I = \int_{0}^{\frac{\pi}{2}} \cos^2 \left( \frac{\pi}{2} - x \right) \, dx = \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx \quad \dots(2) \]

Adding (1) and (2):

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

Solution:

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

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

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

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

Adding (1) and (2):

\[ \begin{aligned} 2I &= \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \\ 2I &= \int_{0}^{\frac{\pi}{2}} 1 \, dx = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} \\ I &= \frac{\pi}{4} \end{aligned} \]

Solution:

This is identical in structure to Question 2. Using the same property \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \), the numerator becomes \( \cos^{3/2} x \).

Adding the two integrals gives \( \int 1 \, dx \).

Result: \( I = \frac{\pi}{4} \).

Solution:

Similar to Questions 2 and 3.

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

Using the property, \( I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^5 x}{\cos^5 x + \sin^5 x} \, dx \).

Adding gives \( 2I = \frac{\pi}{2} \Rightarrow I = \frac{\pi}{4} \).

Solution:

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

Split limits at \( x = -2 \):

\[ \begin{aligned} I &= \int_{-5}^{-2} -(x+2) \, dx + \int_{-2}^{5} (x+2) \, dx \\ &= -\left[ \frac{x^2}{2} + 2x \right]_{-5}^{-2} + \left[ \frac{x^2}{2} + 2x \right]_{-2}^{5} \\ &= -\left[ (2-4) - (\frac{25}{2}-10) \right] + \left[ (\frac{25}{2}+10) - (2-4) \right] \\ &= -\left[ -2 - 2.5 \right] + \left[ 22.5 - (-2) \right] \\ &= -[-4.5] + [24.5] \\ &= 4.5 + 24.5 = 29 \end{aligned} \]

Solution:

Split limits at \( x = 5 \).

\[ \begin{aligned} I &= \int_{2}^{5} -(x-5) \, dx + \int_{5}^{8} (x-5) \, dx \\ &= -\left[ \frac{x^2}{2} - 5x \right]_{2}^{5} + \left[ \frac{x^2}{2} - 5x \right]_{5}^{8} \\ &= -\left[ (\frac{25}{2}-25) - (2-10) \right] + \left[ (32-40) - (\frac{25}{2}-25) \right] \\ &= -[ -12.5 - (-8) ] + [ -8 - (-12.5) ] \\ &= -[-4.5] + [4.5] \\ &= 4.5 + 4.5 = 9 \end{aligned} \]

Solution:

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

\[ \begin{aligned} I &= \int_{0}^{1} (1-x)(1-(1-x))^n \, dx \\ &= \int_{0}^{1} (1-x) x^n \, dx \\ &= \int_{0}^{1} (x^n - x^{n+1}) \, dx \\ &= \left[ \frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2} \right]_{0}^{1} \\ &= \frac{1}{n+1} - \frac{1}{n+2} \\ &= \frac{(n+2)-(n+1)}{(n+1)(n+2)} \\ &= \frac{1}{(n+1)(n+2)} \end{aligned} \]

Solution:

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

\[ \begin{aligned} I &= \int_{0}^{\frac{\pi}{4}} \log\left(1 + \tan\left(\frac{\pi}{4}-x\right)\right) \, dx \\ &= \int_{0}^{\frac{\pi}{4}} \log\left(1 + \frac{1-\tan x}{1+\tan x}\right) \, dx \\ &= \int_{0}^{\frac{\pi}{4}} \log\left(\frac{1+\tan x + 1-\tan x}{1+\tan x}\right) \, dx \\ &= \int_{0}^{\frac{\pi}{4}} \log\left(\frac{2}{1+\tan x}\right) \, dx \\ &= \int_{0}^{\frac{\pi}{4}} (\log 2 - \log(1+\tan x)) \, dx \\ I &= \int_{0}^{\frac{\pi}{4}} \log 2 \, dx - I \\ 2I &= \log 2 [x]_{0}^{\frac{\pi}{4}} = \frac{\pi}{4} \log 2 \\ I &= \frac{\pi}{8} \log 2 \end{aligned} \]

Solution:

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

\[ \begin{aligned} I &= \int_{0}^{2} (2-x)\sqrt{2-(2-x)} \, dx \\ &= \int_{0}^{2} (2-x)\sqrt{x} \, dx \\ &= \int_{0}^{2} (2x^{1/2} - x^{3/2}) \, dx \\ &= \left[ 2 \frac{x^{3/2}}{3/2} - \frac{x^{5/2}}{5/2} \right]_{0}^{2} \\ &= \frac{4}{3}(2)^{3/2} - \frac{2}{5}(2)^{5/2} \\ &= \frac{4}{3}(2\sqrt{2}) - \frac{2}{5}(4\sqrt{2}) \\ &= \frac{8\sqrt{2}}{3} - \frac{8\sqrt{2}}{5} \\ &= 8\sqrt{2} \left( \frac{1}{3} - \frac{1}{5} \right) \\ &= 8\sqrt{2} \left( \frac{2}{15} \right) = \frac{16\sqrt{2}}{15} \end{aligned} \]

Solution:

Rewrite integrand:

\[ \begin{aligned} 2\log \sin x - \log(2\sin x \cos x) &= \log \sin^2 x - (\log 2 + \log \sin x + \log \cos x) \\ &= \log \sin x - \log \cos x - \log 2 \\ &= \log(\tan x) - \log 2 \end{aligned} \]

Thus, \( I = \int_{0}^{\frac{\pi}{2}} \log(\tan x) \, dx - \int_{0}^{\frac{\pi}{2}} \log 2 \, dx \).

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

\[ I = 0 - \log 2 [x]_{0}^{\frac{\pi}{2}} = -\frac{\pi}{2} \log 2 = \frac{\pi}{2} \log \left(\frac{1}{2}\right) \]

Solution:

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

Using property for even functions:

\[ I = 2 \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx \]

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

Therefore, \( I = 2 \left( \frac{\pi}{4} \right) = \frac{\pi}{2} \).

Solution:

Use property \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \).

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

Adding the two expressions for I:

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

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

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

Thus, \( I = \pi \).

Solution:

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

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

Since \( f(x) \) is an odd function, the integral over symmetric limits is 0.

Therefore, \( I = 0 \).

Solution:

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

\( f(2\pi - x) = \cos^5(2\pi - x) = \cos^5 x = f(x) \).

Using property, \( I = 2 \int_{0}^{\pi} \cos^5 x \, dx \).

Now consider the new integral limit \( \pi \). \( f(\pi - x) = \cos^5(\pi - x) = (-\cos x)^5 = -\cos^5 x = -f(x) \).

Using property, if \( f(2a-x) = -f(x) \), then \( \int_{0}^{2a} f(x) dx = 0 \).

Here \( 2a = \pi \). Thus, the integral is 0.

Answer: 0.

Solution:

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

\[ \begin{aligned} I &= \int_{0}^{\frac{\pi}{2}} \frac{\sin(\frac{\pi}{2}-x) - \cos(\frac{\pi}{2}-x)}{1 + \sin(\frac{\pi}{2}-x) \cos(\frac{\pi}{2}-x)} \, dx \\ &= \int_{0}^{\frac{\pi}{2}} \frac{\cos x - \sin x}{1 + \cos x \sin x} \, dx \\ &= - \int_{0}^{\frac{\pi}{2}} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx \\ I &= -I \end{aligned} \]

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

Solution:

Standard integral result.

Let \( I = \int_{0}^{\pi} \log(1 + \cos x) \, dx \). Using P4, \( I = \int_{0}^{\pi} \log(1 - \cos x) \, dx \).

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

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

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

So, \( I = 2(-\frac{\pi}{2} \log 2) = -\pi \log 2 = \pi \log(\frac{1}{2}) \).

Solution:

Similar to Question 2.

Using property P4, the numerator becomes \( \sqrt{a-x} \).

Adding original and transformed integrals:

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

\( I = \frac{a}{2} \).

Solution:

Split limits at \( x=1 \).

\[ \begin{aligned} I &= \int_{0}^{1} -(x-1) \, dx + \int_{1}^{4} (x-1) \, dx \\ &= \left[ x - \frac{x^2}{2} \right]_{0}^{1} + \left[ \frac{x^2}{2} - x \right]_{1}^{4} \\ &= (1 - 0.5) + [(8-4) - (0.5-1)] \\ &= 0.5 + [4 - (-0.5)] \\ &= 0.5 + 4.5 = 5 \end{aligned} \]

Solution:

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

Using property P4: \( I = \int_{0}^{a} f(a-x)g(a-x) \, dx \).

Given \( f(a-x) = f(x) \) and \( g(a-x) = 4 - g(x) \).

\[ \begin{aligned} I &= \int_{0}^{a} f(x)(4 - g(x)) \, dx \\ I &= 4 \int_{0}^{a} f(x) \, dx - \int_{0}^{a} f(x)g(x) \, dx \\ I &= 4 \int_{0}^{a} f(x) \, dx - I \\ 2I &= 4 \int_{0}^{a} f(x) \, dx \\ I &= 2 \int_{0}^{a} f(x) \, dx \end{aligned} \]

Answer: (C)

Solution:

Check functions for odd/even:

• \( f(x) = x^3 \): Odd. Integral = 0.
• \( f(x) = x \cos x \): Odd (since \(-x \cos(-x) = -x \cos x\)). Integral = 0.
• \( f(x) = \tan^5 x \): Odd. Integral = 0.
• \( f(x) = 1 \): Even.

Thus, \( I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx = [x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi \).

Answer: (C)

Solution:

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

Using P4, replace \( x \) with \( \frac{\pi}{2}-x \). Sin becomes Cos, Cos becomes Sin.

\[ \begin{aligned} I &= \int_{0}^{\frac{\pi}{2}} \log \left( \frac{4+3\cos x}{4+3\sin x} \right) \, dx \\ &= \int_{0}^{\frac{\pi}{2}} \log \left( \frac{4+3\sin x}{4+3\cos x} \right)^{-1} \, dx \\ &= - \int_{0}^{\frac{\pi}{2}} \log \left( \frac{4+3\sin x}{4+3\cos x} \right) \, dx \\ I &= -I \Rightarrow 2I = 0 \Rightarrow I = 0 \end{aligned} \]