$\int_0^{\pi\over2}{\cos x\sin x\over(x+1)}dx$ = ${1\over2}({1\over2}+{1\over\pi+2}$- $\int_0^\pi{\cos x\over(x+2)^2}dx)$ $$A=\int_0^\pi{\cos x\over(x+2)^2}dx$$
Then prove that, $\displaystyle\int_0^{\Large\pi\over2}{\cos x\sin x\over(x+1)}dx={1\over2}\left({1\over2}+{1\over\pi+2}-A\right)$
 A: Rewrite:
$$
\int_0^{\Large\pi\over2}{\cos x\sin x\over x+1}dx=\frac12\int_0^{\Large\pi\over2}{\sin 2x\over (x+1)}dx
$$
Now, use IBP by taking $u=\dfrac1{x+1}\;\Rightarrow\;du=-\dfrac{dx}{(x+1)^2}$ and
$$
dv=\sin2x\ dx\quad\Rightarrow\quad v=-\frac12\cos2x
$$
Hence
$$
\begin{align}
\int_0^{\Large\pi\over2}{\sin 2x\over (x+1)}dx
&=-\left.\dfrac{\cos2x}{2(x+1)}\right|_0^{\Large\pi\over2}-\frac12\int_0^{\Large\pi\over2}\dfrac{\cos2x}{(x+1)^2}dx\\
&=\frac{1}{\pi+2}+\frac{1}{2}-\frac12\int_0^{\Large\pi\over2}\dfrac{\cos2x}{(x+1)^2}dx.
\end{align}
$$
Now, let $y=2x\;\Rightarrow\;dx=\dfrac{dy}2$ and $0<x<\dfrac\pi2$ is corresponding to $0<y<\pi$. Thus
$$
\int_0^{\Large\pi\over2}{\cos x\sin x\over x+1}dx=\frac12\left(\frac{1}{\pi+2}+\frac{1}{2}-\int_0^{\Large\pi}\dfrac{\cos y}{(y+2)^2}dy\right).
$$
A: HINT:
Start with $x=2y$ to find $$\int\frac{\cos x}{(2+x)^2}\ dx=\frac12\int\frac{\cos2y}{(1+y)^2}dy$$
Now integrate by parts $\displaystyle\int\frac{\cos2y}{(1+y)^2}dy=\cos2y\int\frac{dy}{(1+y)^2}-\int\left(\frac{d(\cos2y)}{dy}\cdot\int\frac{dy}{(1+y)^2}\right)\ dy$ 
