Question

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

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

Answer 1

Answer: (A)

We have,

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

$=\underset{0}{\overset{\pi /2}{\int }}\frac{co{s}^2\frac{x}{2}-si{n}^2\frac{x}{2}}{{\left(cos\frac{x}{2}+sin\frac{x}{2}\right)}^3}dx=\underset{0}{\overset{\pi /2}{\int }}\frac{co{s}^{}\frac{x}{2}-sin\frac{x}{2}}{{\left(cos\frac{x}{2}+sin\frac{x}{2}\right)}^2}dx$

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 $x=\frac{\pi }{2}\Rightarrow t=\sqrt{2}$

$\therefore I=\underset{0}{\overset{\sqrt{2}}{\int }}\frac{2dt}{t^2}2{\left[-\frac{1}{t}\right]}_1^{\sqrt{2}}$

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