MCQ
Evaluate : $\int_0^{\pi / 2} \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x$
  • A
    $2-\sqrt{2}$
  • B
    $2+\sqrt{2}$
  • C
    $3+\sqrt{3}$
  • D
    $3-\sqrt{3}$

Answer

$\begin{aligned} & \text {(a) : Let, } I=\int_0^{\pi / 2} \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x \\ & =\int_0^{\pi / 2} \frac{\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x=\int_0^{\pi / 2} \frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2} d x\end{aligned}$
Let $\cos \frac{x}{2}+\sin \frac{x}{2}=t \Rightarrow \frac{1}{2}\left(-\sin \frac{x}{2}+\cos \frac{x}{2}\right) d x=d t$
Also, $x=0 \Rightarrow t=1$ and $x=\frac{\pi}{2} \Rightarrow t=\sqrt{2}$
$
\therefore \quad I=\int_1^{\sqrt{2}} \frac{2 d t}{t^2}=2\left[-\frac{1}{t}\right]_1^{\sqrt{2}}=2\left[-\frac{1}{\sqrt{2}}+1\right]=2-\sqrt{2}
$

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