Q128 / 200 · WBJEE 2014 · Mathematics
Suppose $M=\int_{0}^{\pi / 2} \frac{\cos x}{x+2} d x$ $N=\int_{0}^{\pi / 4} \frac{\sin x \cos x}{(x+1)^{2}} d x$ Then, the value of $(M-N)$ equals
- A$\frac{3}{\pi+2}$
- B$\frac{2}{\pi-4}$
- C$\frac{4}{\pi-2}$
- Correct answerD$\frac{2}{\pi+4}$
Solution
Given, $M=\int_{0}^{\pi / 2} \frac{\cos x}{(x+2)} d x$
and $\quad N=\int_{0}^{\sqrt{4}} \frac{\sin x \cos x}{(x+1)^{2}} d x$
$\Rightarrow \quad N=\int_{0}^{\pi / 4} \frac{1}{2} \frac{\sin 2 x}{(x+1)^{2}} d x$
Put $2 x=t \Rightarrow d x=\frac{d t}{2}$
$\begin{aligned} N &=\int_{0}^{\pi / 2} \frac{\sin t}{4\left(\frac{t}{2}+1\right)^{2}} d t \\ &=\int_{0}^{\pi / 2} \frac{\sin t}{(t+2)^{2}} d t \\ &=\int_{0}^{\pi / 2} \frac{\sin x}{(x+2)^{2}} d x \end{aligned}$
$M-N=\int_{0}^{\pi / 2} \cos x \times \frac{1}{(x+2)} d x$
$=\left[\frac{\sin x}{x+2}\right]_{0}^{\pi / 2}-\int_{0}^{\pi / 2}-\frac{\sin x}{(x+2)^{2}} d x$
$-\int_{0}^{\pi / 2} \frac{\sin x}{(x+2)^{2}} d x$
$=\frac{\sin \frac{\pi}{2}}{\frac{\pi}{2}+2}=\frac{1}{\frac{\pi+4}{2}}=\frac{2}{\pi+4}$
Hence, the correct answer is option: 'D'.