Question

If $M={\int }_0^{\pi /2}\frac{\cos x}{x+2}dx,N={\int }_0^{\pi /4}\frac{\sin x\cos x}{{(x+1)}^2}dx$, then the value of $M-N$ is ?

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

Answer 1

Answer: (D)

$\frac{2}{\pi +4}$

$N={\int }_0^{\pi /4}\frac{\sin 2x}{2{(x+1)}^2}dx$

Let $2x=t$

$2dx=dt$

$N={\int }_0^{\pi /2}\frac{\sin t}{4(\frac{t}{2}+1)}dt={\int }_0^{\pi /2}\frac{\sin t}{{(t+2)}^2}dt$

$\sin t(-\frac{1}{t+2})+{\int }_0^{\pi /2}\frac{\cos t}{{(t+2)}^{}}dt$

$N={(-\sin t\left(\frac{1}{t+2}\right))}_0^{\pi /2}+M$

$M-N=\frac{2}{\pi +4}$