• # question_answer Let $M=\int\limits_{0}^{\frac{\pi }{2}}{\frac{\cos x}{x+2}dx}$ and $N=\int\limits_{0}^{\frac{\pi }{4}}{\frac{\sin x.\cos x}{{{(x+1)}^{2}}}dx}$ then the value of (M - N) equals A) $\frac{2}{\pi +1}$                                B) $\frac{2}{\pi +2}$ C) $\frac{2}{\pi +4}$ D)  0

$N=\frac{1}{2}\,\int\limits_{0}^{\frac{\pi }{4}}{\frac{\sin \,2x}{{{(x+1)}^{2}}}dx}$ On putting $2x=1,$ we get $N=\frac{1}{4}.\,4\int\limits_{0}^{\frac{\pi }{2}}{\,\frac{\sin t}{{{(t+2)}^{2}}}dt}$ $\left. N=(\sin t)\,\left( \frac{-1}{t+2} \right) \right]_{0}^{\frac{\pi }{2}}+\int\limits_{0}^{\frac{\pi }{2}}{\frac{\cos t}{t+2}dt}$ $\left. \therefore \,\,M-N\,=\frac{\sin t}{\operatorname{t}+2} \right]_{0}^{\frac{\pi }{2}}\,=\frac{2}{\pi +4}$