If $A=\int\limits_0^\pi \frac{\cos x}{(x+2)^2} d x$, then $\int\limits_0^{\frac{\pi}{2}} \frac{\sin 2 x}{(x+1)} d x$ is equal to
Answer & explanation
Correct answer: option 4
Given $A=\int\limits_0^\pi \frac{\cos x}{(x+2)^2} d x$
$A=\left(\frac{-\cos x}{x+2}\right)_0^\pi-\int\limits_0^\pi \frac{\sin x}{(x+2)} d x \Rightarrow \int\limits_0^\pi \frac{\sin x}{(x+2)} d x=\frac{1}{2}+\frac{1}{\pi+2}-A$
Now let $I = \int\limits_0^\pi \frac{\sin x}{(x+2)} d x$ put $x=2 t \Rightarrow dx=2 dt \Rightarrow I=\int\limits_0^{\pi / 2} \frac{\sin 2 t}{(t+1)} d t=\int\limits_0^{\pi / 2} \frac{\sin 2 x}{(x+1)} d x$
$\Rightarrow I=\frac{1}{2}+\frac{1}{\pi+2}-A$