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If \(M=\int_{0}^{\pi / 2} \frac{\cos x}{x+2} d x, N=\int_{0}^{\pi / 4} \frac{\sin x \cos x}{(x+1)^{2}} d x\), then the value of \(M-N\) is
(A) \(\pi\)
(B) \(\frac{\pi}{4}\)
(C) \(\frac{2}{\pi-4}\)
(D) \(\frac{2}{\pi+4}\)

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Ans: (D)
Hint \(: N=\int_{0}^{\pi / 4} \frac{\sin 2 x d x}{2(x+1)^{2}}\)
let \(2 x=t, \quad d t=2 d x, \quad N=\int_{0}^{\pi / 2} \frac{\sin t \frac{d t}{2}}{2 \frac{(t+2)^{2}}{4}} \Rightarrow N=\int_{0}^{\pi / 2} \frac{\sin t d t}{(t+2)^{2}}=\left(\frac{-\sin t}{t+2}\right)_{0}^{\pi / 2}+\int_{0}^{\pi / 2} \frac{\cos t}{(t+2)} d t\)
$$
\Rightarrow \mathrm{N}=\frac{-2}{\pi+4}+\mathrm{M} \Rightarrow \mathrm{M}-\mathrm{N}=\frac{2}{\pi+4}
$$
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