Question

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

Given : $M={\int }_0^{\frac{\pi }{2}}\frac{\cos x}{(x+2)}dxN={\int }_0^{\frac{\pi }{4}}\frac{\sin x\cos x}{(x+1{)}^2}dx$

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

Put $2x=t$

$2dx=dt=\frac{1}{2}{\int }_0^{,\frac{\pi }{2}}\frac{4\sin t}{(t+2{)}^2}.\frac{dt}{2}={\int }_0^{,\frac{\pi }{2}}\frac{\sin t}{(t+2{)}^2}.dtM-N={\int }_0^{,\frac{\pi }{2}}[\frac{\cos x}{(x+2)}-\frac{\sin x}{(x+2{)}^2}]dx={\int }_0^{\frac{\pi }{2}}\frac{\cos x}{(x+2)}-{\int }_0^{,\frac{\pi }{2}}\frac{\sin t}{(x+2{)}^2}M-N={[\frac{1}{(x+2)}.\sin x]}_0^{\frac{\pi }{2}}+{\int }_0^{,\frac{\pi }{2}}\frac{\sin x}{(x+2{)}^2}-{\int }_0^{,\frac{\pi }{2}}\frac{\sin x}{(x+2{)}^2}M-N=\frac{2}{\pi +4}$

Hence the correct answer is $\frac{2}{\pi +4}$