Question

Match the elements of List I with List II:

List-I	List-II
A) ${\int }_0^{\frac{\pi }{2}}\log \sin xdx=$	1) $4\pi$
B) ${\int }_0^{\frac{\pi }{2}}\log \tan xdx=$	2) $-\frac{\pi }{2}{\log }_{e}2$
C) ${\int }_0^{\pi }x\log \sin xdx=$	3) $-\frac{{\pi }^2}{2}{\log }_{e}2$
D) ${\int }_{-\pi }^{\pi }(x^3+x\cos x+{\tan }^{5}x+2)dx=$	4) $0$

• A. $A-3,B-4,C-1,D-2$
• B. $A-1,B-4,C-2,D-3$
• C. $A-3,B-2,C-2,D-1$
• D. $A-2,B-4,C-3,D-1$

Answer 1

Answer: (D)

$A-2,B-4,C-3,D-1$

A) Let, $I={\int }_0^{\frac{\pi }{2}}\log \sin xdx$ ..(1)

Using integration property, replace $x$ with $\frac{\pi }{2}-x$, we get
$I={\int }_0^{\frac{\pi }{2}}\log \cos xdx$ ..(2)

Adding (1) and (2), we get
$\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}\log \left(\sin x\cos x\right)dx$

$\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}(\log \sin 2x-\log 2)dx$
$\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}\log \sin 2xdx-{\int }_0^{\frac{\pi }{2}}\log 2dx$

Substitute, $2x=t\Rightarrow 2dx=dt$

When, $x=0\Rightarrow t=0$; $x=\frac{\pi }{2}\Rightarrow t=\pi$

Therefore,

$\Rightarrow 2I=\frac{1}{2}{\int }_0^{\pi }\log \sin tdt-\frac{\pi }{2}\log 2$
$\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}\log \sin tdt-\frac{\pi }{2}\log 2$

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

$\Rightarrow I=-\frac{\pi }{2}\log 2$
B) Let $I={\int }_0^{\frac{\pi }{2}}\log \tan xdx$
$I={\int }_0^{\frac{\pi }{2}}\log \cot xdx$
$2I={\int }_0^{\frac{\pi }{2}}\log \cot xdx+{\int }_0^{\frac{\pi }{2}}\log \tan xdx$

$2I={\int }_0^{\frac{\pi }{2}}\log \left(\tan x\cot x\right)dx$

$\Rightarrow I=0$ ($\because \log 1=0$)

C) Let, $I={\int }_0^{\pi }x\log \sin xdx$
$I={\int }_0^{\pi }(\pi -x)\log \sin xdx$ ($\because \sin (\pi -x)=\sin x$)
$\Rightarrow 2I=\pi {\int }_0^{\pi }\log \sin xdx$
$\Rightarrow 2I=2\pi {\int }_0^{\frac{\pi }{2}}\log \sin xdx$
$I=\pi (-\frac{\pi }{2}{\log }_{e}2)$

$(\because {\int }_0^{\frac{\pi }{2}}\log \sin xdx=-\frac{\pi }{2}{\log }_{e}2)$

$I=-\frac{{\pi }^2}{2}lo{g}_{e}2$

D) $I={\int }_{-\pi }^{\pi }(x^3+x\cos x+{\tan }^{5}x+2)dx$

Replace $x$ by $-x$, we get

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

Adding both, we get

$2I={\int }_{-\pi }^{\pi }\left(4\right)dx$

$I=4\pi$