Question

Evaluate : $ \int\limits_{0}^{\pi/2}\frac{ cos \,x}{\left(cos \frac{x}{2} +sin \frac{x}{2}\right)^{3}} dx$

• A. $2 -\sqrt{2}$
• B. $2 +\sqrt{2}$
• C. $3 +\sqrt{3}$
• D. $3 -\sqrt{3}$

Answer 1

Answer: (A)

We have,

$I = \int\limits_{0}^{\pi/2}\frac{ cos \,x}{\left(cos \frac{x}{2} +sin \frac{x}{2}\right)^{3}} dx $

$= \int\limits_{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}} dx = \int\limits_{0}^{\pi /2} \frac{cos^{ } \frac{x}{2} - sin \frac{x}{2} }{\left(cos \frac{x}{2} + sin \frac{x}{2}\right)^{2}} dx $

Let $cos \frac{x}{2} +sin \frac{x}{2} = t$
$ \Rightarrow \frac{1}{2}\left(-sin \frac{x}{2} +cos \frac{x}{2}\right) dx = dt $

Also, $x = 0 \Rightarrow t = 1\,$ and $\, x = \frac{\pi}{2} \Rightarrow t = \sqrt{2}$

$\therefore I = \int\limits_{0}^{\sqrt{2}} \frac{2dt}{t^{2}} 2\left[-\frac{1}{t}\right]_{1}^{\sqrt{2}} $

$= 2\left[-\frac{1}{\sqrt{2}} +1\right] = \left(2-\sqrt{2}\right)$