integration

If A = $\int_{0}^{\pi}\frac{cosx}{(x+2)^2}dx$ , then show that $\int_{0}^{\pi /2}\frac{sinxcosx}{x+1}dx$ = $\frac{1}{2}$($\frac{1}{2}$+$\frac{1}{\pi +2}-A $)

Solution :Let $u = \cos x\Rightarrow du = -\sin x\,dx$ and $dv = \frac{1}{\left(x+2\right)^{2}}\,dx\Rightarrow v = -\frac{1}{x+2}$. Integration by parts method,
\begin{align*}
\int_{0}^{\pi}\frac{\cos x}{\left(x+2\right)^{2}}\,dx & = -\frac{\cos x}{x+2}\bigg]_{0}^{\pi}-\int_{0}^{\pi}\frac{\sin x}{x+2}\,dx\\
 \int_{0}^{\pi}\frac{\sin x}{x+2}\,dx & = \left(\frac{1}{\pi+2}+\frac{1}{2}\right) - A
\end{align*}Now let $ x=2a\Rightarrow dx = 2\,da$.
\begin{align*}
 \int_{0}^{\frac{\pi}{2}}\frac{\sin 2a}{2a+2}2\,da & = \left(\frac{1}{\pi+2}+\frac{1}{2}\right) - A\\
2\int_{0}^{\frac{\pi}{2}}\frac{\sin x\cos x}{x+1}\,dx & = \left(\frac{1}{\pi+2}+\frac{1}{2}\right) - A\\
\int_{0}^{\frac{\pi}{2}}\frac{\sin x\cos x}{x+1}\,dx & = \frac{1}{2}\left(\frac{1}{\pi+2}+\frac{1}{2}-A\right)
\end{align*}

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