Question

Suppose $M=\int\limits_{0}^{\pi / 2} \frac{\cos x}{x+2} d x$, $N=\int\limits_{0}^{\pi / 4} \frac{\sin x \cos x}{(x+1)^{2}} d x$ Then, the value of $(M-N)$ equals

• A. $\frac{3}{\pi+2}$
• B. $\frac{2}{\pi-4}$
• C. $\frac{4}{\pi-2}$
• D. $\frac{2}{\pi+4}$

Answer 1

The Correct Option is D

Given, $M=\int\limits_{0}^{\pi / 2} \frac{\cos x}{(x+2)} d x $
and $ N=\int\limits_{0}^{\pi / 4} \frac{\sin x \cos x}{(x+1)^{2}} d x $
$ \Rightarrow N=\int\limits_{0}^{\pi / 4} \frac{1}{2} \frac{\sin 2 x}{(x+1)^{2}} d x $
Put$ 2 x =t $
$\Rightarrow d x=\frac{d t}{2}$
$\therefore N=\int\limits_{0}^{\pi / 2} \frac{\sin t}{4\left(\frac{t}{2}+1\right)^{2}} d t$
$=\int\limits_{0}^{\pi / 2} \frac{\sin t}{(t+2)^{2}} d t$
$=\int\limits_{0}^{\pi / 2} \frac{\sin x}{(x+2)^{2}} d x$
$\therefore M-N=\int\limits_{0}^{\pi / 2} \underset{II}{\cos x} \times \underset{I}{\frac{1}{(x+2)}} d x -\int\limits_{0}^{\pi / 2} \frac{\sin x}{(x+2)^{2}} d x$
$=\left[\frac{\sin x}{x+2}\right]_{0}^{\pi / 2}-\int\limits_{0}^{\pi / 2}-\frac{\sin x}{(x+2)^{2}} d x -\int\limits_{0}^{\pi / 2} \frac{\sin x}{(x+2)^{2}} d x$
$=\frac{\sin \frac{\pi}{2}}{\frac{\pi}{2}+2}=\frac{1}{\frac{\pi+4}{2}}=\frac{2}{\pi+4}$