Solution

$I=\int_0^{\dfrac{\pi}{2}} \cos ^{2} x d x \qquad…(1) $

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \cos ^{2}(\dfrac{\pi}{2}-x) d x \qquad (\because\int _0^{a} f(x) d x=\int _0^{a} f(a-x) d x)$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \sin ^{2} x d x \qquad…(2)$

Adding (1) and (2), we obtain

$2 I=\int_0^{\dfrac{\pi}{2}}(\sin ^{2} x+\cos ^{2} x) d x$

$\Rightarrow 2 I=\int_0^{\dfrac{\pi}{2}} 1 \cdot d x$

$\Rightarrow 2 I=[x]_0^{\dfrac{\pi}{2}}$

$\Rightarrow 2 I=\dfrac{\pi}{2}$

$\Rightarrow I=\dfrac{\pi}{4}$

Solution

$\int_0^{\dfrac{\pi}{2}} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$

Let $I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \qquad…(1)$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sqrt{\sin (\dfrac{\pi}{2}-x)}}{\sqrt{\sin (\dfrac{\pi}{2}-x)}+\sqrt{\cos (\dfrac{\pi}{2}-x)}} d x $

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sqrt{\cos(x) }}{\sqrt{\cos(x) }+\sqrt{\sin x}} d x \qquad…(2)$

Adding (1) and (2), we obtain

$2 I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$

$\Rightarrow 2 I=\int_0^{\dfrac{\pi}{2}} 1 \cdot d x$

$\Rightarrow 2 I=[x]_0^{\dfrac{\pi}{2}}$

$\Rightarrow 2 I=\dfrac{\pi}{2}$

$\Rightarrow I=\dfrac{\pi}{4}$

Solution

$(\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x)$

Let $I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin ^{\dfrac{3}{2}} x}{\sin ^{\dfrac{3}{2}} x+\cos ^{\dfrac{3}{2}} x} d x \qquad…(1)$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin ^{\dfrac{3}{2}}(\dfrac{\pi}{2}-x)}{\sin ^{\dfrac{3}{2}}(\dfrac{\pi}{2}-x)+\cos ^{\dfrac{3}{2}}(\dfrac{\pi}{2}-x)} d x$

$ (\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x) $

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \dfrac{\cos ^{\dfrac{3}{2}} x}{\sin ^{\dfrac{3}{2}} x+\cos ^{\dfrac{3}{2}} x} d x \qquad…(2)$

Adding (1) and (2), we obtain

$2 I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin ^{\dfrac{3}{2}} x+\cos ^{\dfrac{3}{2}} x}{\sin ^{\dfrac{3}{2}} x+\cos ^{\dfrac{3}{2}} x} d x$

$\Rightarrow 2 I=\int_0^{\dfrac{\pi}{2}} 1 \cdot d x$

$\Rightarrow 2 I=[x]_0^{\dfrac{\pi}{2}}$

$\Rightarrow 2 I=\dfrac{\pi}{2}$

$\Rightarrow I=\dfrac{\pi}{4}$

Solution

Let $I=\int_0^{\dfrac{\pi}{2}} \dfrac{\cos ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x \qquad…(1)$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \dfrac{\cos ^{5}(\dfrac{\pi}{2}-x)}{\sin ^{5}(\dfrac{\pi}{2}-x)+\cos ^{5}(\dfrac{\pi}{2}-x)} d x $

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x \qquad…(2)$

Adding (1) and (2), we obtain

$2 I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin ^{5} x+\cos ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x$

$\Rightarrow 2 I=\int_0^{\dfrac{\pi}{2}} 1 \cdot d x$

$\Rightarrow 2 I=[x]_0^{\dfrac{\pi}{2}}$

$\Rightarrow 2 I=\dfrac{\pi}{2}$

$\Rightarrow I=\dfrac{\pi}{4}$

Solution

Let $I=\int _{-5}^{5}|x+2| d x$

It can be seen that $(x+2) \leq 0$ on $[-5,-2]$ and $(x+2) \geq 0$ on $[-2,5]$.

$ \begin{aligned} \therefore I & =\int _{-5}^{-2}-(x+2) d x+\int _{-2}^{5}(x+2) d x \quad(\because\int_a^{b} f(x)=\int_a^{c} f(x)+\int_c^{b} f(x)) \\ I & =-[\dfrac{x^{2}}{2}+2 x] _{-5}^{-2}+[\dfrac{x^{2}}{2}+2 x] _{-2}^{5} \\ & =-[\dfrac{(-2)^{2}}{2}+2(-2)-\dfrac{(-5)^{2}}{2}-2(-5)]+[\dfrac{(5)^{2}}{2}+2(5)-\dfrac{(-2)^{2}}{2}-2(-2)] \\ & =-[2-4-\dfrac{25}{2}+10]+[\dfrac{25}{2}+10-2+4] \\ & =-2+4+\dfrac{25}{2}-10+\dfrac{25}{2}+10-2+4 \\ & =29 \end{aligned} $

Solution

Let $I=\int_2^{8}|x-5| d x$

It can be seen that $(x-5) \leq 0$ on $[2,5]$ and $(x-5) \geq 0$ on $[5,8]$.

$ \begin{aligned} &I=\int_2^{5}-(x-5) d x+\int_5^{8}(x-5) d x & (\because\int_a^{b} f(x)=\int_a^{c} f(x)+\int_c^{b} f(x)) \ & =-[\dfrac{x^{2}}{2}-5 x]_2^{5}+[\dfrac{x^{2}}{2}-5 x]_5^{8} \ & =-[\dfrac{25}{2}-25-2+10]+[32-40-\dfrac{25}{2}+25] \ &=9 \end{aligned} $

Solution

$ \begin{aligned} \text{ Let } & I=\int_0^{1} x(1-x)^{n} d x \\ \therefore I & =\int_0^{1}(1-x)(1-(1-x))^{n} d x \\ & =\int_0^{1}(1-x)(x)^{n} d x \\ & =\int_0^{1}(x^{n}-x^{n+1}) d x \\ & =\bigg[\dfrac{x^{n+1}}{n+1}-\dfrac{x^{n+2}}{n+2}\bigg]_0^{1} \qquad(\because\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x) \\ & =\bigg[\dfrac{1}{n+1}-\dfrac{1}{n+2}\bigg] \\ & =\dfrac{(n+2)-(n+1)}{(n+1)(n+2)} \\ & =\dfrac{1}{(n+1)(n+2)} \end{aligned} $

Solution

Let $I=\int_0^{\dfrac{\pi}{4}} \log (1+\tan x) d x$

$\therefore I=\int_0^{\dfrac{\pi}{4}} \log [1+\tan (\dfrac{\pi}{4}-x)] d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{4}} \log\bigg(1+\dfrac{\tan \dfrac{\pi}{4}-\tan x}{1+\tan \dfrac{\pi}{4} \tan x}\bigg) d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{4}} \log ({1+\dfrac{1-\tan x}{1+\tan x}}) d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{4}} \log \dfrac{2}{(1+\tan x)} d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{4}} \log 2 d x-\int_0^{\dfrac{\pi}{4}} \log (1+\tan x) d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{4}} \log 2 d x-I$

$\Rightarrow 2 I=[x \log 2]_0^{\dfrac{\pi}{4}}$

$\Rightarrow 2 I=\dfrac{\pi}{4} \log 2$

$\Rightarrow I=\dfrac{\pi}{8} \log 2$

Solution

It is known that, $\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x$

Let $I=\int_0^{2} x \sqrt{2-x} d x$

$ \begin{aligned} I & =\int_0^{2}(2-x) \sqrt{x} d x \\ & =\int_0^{2}{2 x^{\dfrac{1}{2}}-x^{\dfrac{3}{2}}} d x \\ & =\bigg[2(\dfrac{x^{\dfrac{3}{2}}}{\dfrac{3}{2}})-\dfrac{x^{\dfrac{5}{2}}}{\dfrac{5}{2}}\bigg]_0^{2} \\ & =\bigg[\dfrac{4}{3} x^{\dfrac{3}{2}}-\dfrac{2}{5} x^{\dfrac{5}{2}}\bigg]_0^{2} \\ & =\dfrac{4}{3}(2)^{\dfrac{3}{2}}-\dfrac{2}{5}(2)^{\dfrac{5}{2}} \\ & =\dfrac{4 \times 2 \sqrt{2}}{3}-\dfrac{2}{5} \times 4 \sqrt{2} \\ & =\dfrac{8 \sqrt{2}}{3}-\dfrac{8 \sqrt{2}}{5} \\ & =\dfrac{40 \sqrt{2}-24 \sqrt{2}}{15} \\ & =\dfrac{16 \sqrt{2}}{15} \end{aligned} $

Solution

Let $I=\int_0^{\dfrac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x \qquad…(1)$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}}{2 \log \sin x-\log (2 \sin x \cos x)} d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}}{2 \log \sin x-\log \sin x-\log \cos x-\log 2} d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}}{\log \sin x-\log \cos x-\log 2} d x$

It is known that, $(\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x)$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}}{\log \cos x-\log \sin x-\log 2} d x \qquad…(2)$

Adding (1) and (2), we obtain

$ \begin{aligned} & 2 I=\int_0^{\dfrac{\pi}{2}}(-\log 2-\log 2) d x \\ & \Rightarrow 2 I=-2 \log 2 \int_0^{\dfrac{\pi}{2}} 1 d x \\ & \Rightarrow I=-\log 2[\dfrac{\pi}{2}] \\ & \Rightarrow I=\dfrac{\pi}{2}(-\log 2) \\ & \Rightarrow I=\dfrac{\pi}{2}[\log \dfrac{1}{2}] \\ & \Rightarrow I=\dfrac{\pi}{2} \log \dfrac{1}{2} \end{aligned} $

Solution

Let $I=\int _{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} \sin ^{2} x d x$

As $\sin ^{2}(-x)=(\sin (-x))^{2}=(-\sin x)^{2}=\sin ^{2} x$, therefore, $\sin ^{2} x$ is an even function.

It is known that if $f(x)$ is an even function, then $\int _{-a}^{a} f(x) d x=2 \int_0^{a} f(x) d x$

$ \begin{aligned} I & =2 \int_0^{\dfrac{\pi}{2}} \sin ^{2} x d x \\ & =2 \int_0^{\dfrac{\pi}{2}} \dfrac{1-\cos 2 x}{2} d x \\ & =\int_0^{\dfrac{\pi}{2}}(1-\cos 2 x) d x \\ & =[x-\dfrac{\sin 2 x}{2}]_0^{\dfrac{\pi}{2}} \\ & =\dfrac{\pi}{2} \end{aligned} $

Solution

Let $I=\int_0^{\pi} \dfrac{x d x}{1+\sin x} \qquad…(1)$

$\Rightarrow I=\int_0^{\pi} \dfrac{(\pi-x)}{1+\sin (\pi-x)} d x$

$(\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x)$

$\Rightarrow I=\int_0^{\pi} \dfrac{(\pi-x)}{1+\sin x} d x \qquad…(2)$

Adding (1) and (2), we obtain

$ \begin{aligned} & 2 I=\int_0^{\pi} \dfrac{\pi}{1+\sin x} d x \\ & \Rightarrow 2 I=\pi \int_0^{\pi} \dfrac{(1-\sin x)}{(1+\sin x)(1-\sin x)} d x \\ & \Rightarrow 2 I=\pi \int_0^{\pi} \dfrac{1-\sin x}{\cos ^{2} x} d x \\ & \Rightarrow 2 I=\pi \int_0^{\pi}{\sec ^{2} x-\tan x \sec x} d x \\ & \Rightarrow 2 I=\pi[\tan x-\sec x]_0^{\pi} \\ & \Rightarrow 2 I=\pi[2] \\ & \Rightarrow I=\pi \end{aligned} $

Solution

Let $I=\int _{\dfrac{-\pi}{2}}^{\dfrac{\pi}{2}} \sin ^{7} x d x$

As $\sin ^{7}(-x)=(\sin (-x))^{7}=(-\sin x)^{7}=-\sin ^{7} x$, therefore, $\sin ^{2} x$ is an odd function.

It is known that, if $f(x)$ is an odd function, then $\int _{-a}^{a} f(x) d x=0$

$\therefore I=\int _{\dfrac{-\pi}{2}}^{\dfrac{\pi}{2}} \sin ^{7} x d x=0$

Solution

Let $I=\int_0^{2 \pi} \cos ^{5} x d x$

$\cos ^{5}(2 \pi-x)=\cos ^{5} x$

It is known that,

$ \begin{aligned} \int_0^{2 a} f(x) d x & =2 \int_0^{a} f(x) d x \text{, if } f(2 a-x)=f(x) \\ & =0 \text{ if } f(2 a-x)=-f(x) \end{aligned} $

$\therefore I=2 \int_0^{\pi} \cos ^{5} x d x$

$\Rightarrow I=2(0)=0 \quad[\because\cos ^{5}(\pi-x)=-\cos ^{5} x]$

Solution

Let $I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin x-\cos x}{1+\sin x \cos x} d x \qquad…(1)$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin (\dfrac{\pi}{2}-x)-\cos (\dfrac{\pi}{2}-x)}{1+\sin (\dfrac{\pi}{2}-x) \cos (\dfrac{\pi}{2}-x)} d x \qquad (\because\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x) $

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \dfrac{\cos x-\sin x}{1+\sin x \cos x} d x \qquad…(2)$

Adding (1) and (2), we obtain

$ \begin{aligned} & 2 I=\int_0^{\dfrac{\pi}{2}} \dfrac{0}{1+\sin x \cos x} d x \\ & \Rightarrow I=0 \end{aligned} $

Solution

$$ \begin{align*} & \text{ Let } I=\int_0^{\pi} \log (1+\cos x) d x \qquad…(1)\\ & \Rightarrow I=\int_0^{\pi} \log (1+\cos (\pi-x)) d x \\ & \Rightarrow I=\int_0^{\pi} \log (1-\cos x) d x \qquad(\because\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x) \qquad…(2) \end{align*} $$

Adding (1) and (2), we obtain

$2 I=\int_0^{\pi}{\log (1+\cos x)+\log (1-\cos x)} d x$

$\Rightarrow 2 I=\int_0^{\pi} \log (1-\cos ^{2} x) d x$

$\Rightarrow 2 I=\int_0^{\pi} \log \sin ^{2} x d x$

$\Rightarrow 2 I=2 \int_0^{\pi} \log \sin x d x$

$\Rightarrow I=\int_0^{\pi} \log \sin x d x$

$\sin (\pi-x)=\sin x$

$\therefore I=2 \int_0^{\dfrac{\pi}{2}} \log \sin x d x \qquad…(3)$

$\Rightarrow I=2 \int_0^{\dfrac{\pi}{2}} \log \sin (\dfrac{\pi}{2}-x) d x=2 \int_0^{\dfrac{\pi}{2}} \log \cos x d x \qquad…(4)$

Adding (3) and (4), we obtain

$2 I=2 \int_0^{\dfrac{\pi}{2}}(\log \sin x+\log \cos x) d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}}(\log \sin x+\log \cos x+\log 2-\log 2) d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}}(\log 2 \sin x \cos x-\log 2) d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \log \sin 2 x d x-\int_0^{\dfrac{\pi}{2}} \log 2 d x$

Let $2 x=t \Rightarrow 2 d x=d t$

When $x=0, t=0$ and when $x=\dfrac{\pi}{2}, \pi=t$

$\therefore I=\dfrac{1}{2} \int_0^{\pi} \log (\sin t) d t-\dfrac{\pi}{2} \log 2$

$\Rightarrow I=\dfrac{1}{2} I-\dfrac{\pi}{2} \log 2$

$\Rightarrow \dfrac{I}{2}=-\dfrac{\pi}{2} \log 2$

$\Rightarrow I=-\pi \log 2$

Solution

Let $I=\int_0^{a} \dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x \qquad…(1)$

It is known that, $(\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x)$

$I=\int_0^{a} \dfrac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}} d x \qquad…(2)$

Adding (1) and (2), we obtain

$2 I=\int_0^{a} \dfrac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}} d x$

$\Rightarrow 2 I=\int_0^{a} 1 d x$

$\Rightarrow 2 I=[x]_0^{a}$

$\Rightarrow 2 I=a$

$\Rightarrow I=\dfrac{a}{2}$

Solution

$I=\int_0^{4}|x-1| d x$

It can be seen that, $(x-1) \leq 0$ when $0 \leq x \leq 1$ and $(x-1) \geq 0$ when $1 \leq x \leq 4$

$ \begin{aligned} I & =\int_0^{1}|x-1| d x+\int_0^{4}|x-1| d x \qquad(\because\int_a^{b} f(x)=\int_a^{c} f(x)+\int_a^{b} f(x)) \\ & =\int_0^{1}-(x-1) d x+\int_1^{4}(x-1) d x \\ & =[x-\dfrac{x^{2}}{2}]_0^{1}+[\dfrac{x^{2}}{2}-x]_1^{4} \\ & =1-\dfrac{1}{2}+\dfrac{(4)^{2}}{2}-4-\dfrac{1}{2}+1 \\ & =1-\dfrac{1}{2}+8-4-\dfrac{1}{2}+1 \\ & =5 \end{aligned} $

Solution

$$ \begin{align*} & \text{ Let } I=\int_0^{a} f(x) g(x) d x \qquad…(1)\\ & \Rightarrow I=\int_0^{a} f(a-x) g(a-x) d x \\ & \Rightarrow I=\int_0^{a} f(x) g(a-x) d x \qquad…(2) \end{align*} $$

Adding (1) and (2), we obtain

$ \begin{aligned} & 2 I=\int_0^{a}{f(x) g(x)+f(x) g(a-x)} d x \\ & \Rightarrow 2 I=\int_0^{a} [f(x){g(x)+g(a-x)}] d x \\ & \Rightarrow 2 I=\int_0^{a} f(x) \times 4 d x \quad[\because g(x)+g(a-x)=4] \\ & \Rightarrow I=2 \int_0^{a} f(x) d x \end{aligned} $

Solution

Let $I=\int _{\dfrac{-\pi}{2}}^{\dfrac{\pi}{2}}(x^{3}+x \cos x+\tan ^{5} x+1) d x$

$\Rightarrow I=\int _{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} x^{3} d x+\int _{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}}x \cos x+\int _{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} \tan ^{5} x d x+\int _{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} 1 \cdot d x$

It is known that if $f(x)$ is an even function, then $\int _{-a}^{a} f(x) d x=2 \int_0^{a} f(x) d x$ and

if $f(x)$ is an odd function, then $\int _{-a}^{a} f(x) d x=0$

$ \begin{aligned} I & =0+0+0+2 \int_0^{\dfrac{\pi}{2}} 1 \cdot d x \\ & =2[x]_0^{\dfrac{\pi}{2}} \\ & =\dfrac{2 \pi}{2} \\ I & =\pi \end{aligned} $

Hence, the correct Answer is C.

Solution

Let $I=\int_0^{\dfrac{\pi}{2}} \log (\dfrac{4+3 \sin x}{4+3 \cos x}) d x \qquad…(1)$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \log [\dfrac{4+3 \sin (\dfrac{\pi}{2}-x)}{4+3 \cos (\dfrac{\pi}{2}-x)}] d x \quad(\because\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x)$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \log (\dfrac{4+3 \cos x}{4+3 \sin x}) d x \qquad…(2)$

Adding (1) and (2), we obtain

$2 I=\int_0^{\dfrac{\pi}{2}}{\log (\dfrac{4+3 \sin x}{4+3 \cos x})+\log (\dfrac{4+3 \cos x}{4+3 \sin x})} d x$

$\Rightarrow 2 I=\int_0^{\dfrac{\pi}{2}} \log (\dfrac{4+3 \sin x}{4+3 \cos x} \times \dfrac{4+3 \cos x}{4+3 \sin x}) d x$

$\Rightarrow 2 I=\int_0^{\dfrac{\pi}{2}} \log 1 d x$

$\Rightarrow 2 I=\int_0^{\dfrac{\pi}{2}} 0 d x$

$\Rightarrow I=0$

Hence, the correct Answer is C.