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If $\displaystyle{M=\int_{0}^{\pi} \frac{\cos x}{(x+2)^2}~dx}$, then show that $$\int_{0}^{\pi/2}\frac{\sin x \cos x}{x+1}~dx=\frac{1}{2}\left(\frac{1}{2}+\frac{1}{\pi +2}-M\right)$$

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Can you check the expressions you've typed in? I'm not convinced that any of the four options given is correct. – Chris Taylor Jan 2 '13 at 10:32
I copy the question from a question set.I will edit my question when I get information about the question. – Argha Jan 2 '13 at 11:56
I think previous question has typo mistake.This question is not completely similar to the previous but more or less similar type.This is copied from "Test of Mathematics at 10+2 level".I am extremely sorry for the mistake.This mistake was not willful. – Argha Jan 2 '13 at 12:30
up vote 3 down vote accepted

$M-N$ comes out to $$ \left(\mathrm{Ci}\left(2+\frac\pi2\right)-\mathrm{Ci}(10)\right)\cos(2)+\frac15\sin(2)\left(5\mathrm{Si}\left(2+\frac\pi2\right)-5 \mathrm{Si}(10)+\cos(2)+\cos(6)\right) $$ where $\mathrm{Si}(x)$ and $\mathrm{Ci}(x)$ are the Sine and Cosine Integrals.

Evaluating this in Mathematica yields $0.25302541534920158619$, which is none of the answers given.

Answer to the current version of the question: $$ \begin{align} \int_0^{\pi/2}\frac{\sin(x)\cos(x)}{x+1}\,\mathrm{d}x &=\int_0^{\pi/2}\frac{\sin(2x)}{2x+2}\,\mathrm{d}x\\ &=\frac12\int_0^\pi\frac{\sin(y)}{y+2}\,\mathrm{d}y\\ &=\left.-\frac12\frac{\cos(y)}{y+2}\right]_0^\pi-\frac12\int_0^\pi\frac{\cos(y)}{(y+2)^2}\,\mathrm{d}y\\ &=\frac12\left(\frac12+\frac1{\pi+2}-\int_0^\pi\frac{\cos(x)}{(x+2)^2}\,\mathrm{d}x\right) \end{align} $$

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thanks for your answer.I also belief that there are some mistake. I edit my question. Please see it. – Argha Jan 2 '13 at 12:34

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