Question

${\int }_0^{\pi /2}\frac{\sin x}{\sqrt{1+\mathrm{cosx}}}dx=$

• A. $\sqrt{2}-1$
• B. $2\sqrt{2}$
• C. $2(\sqrt{2}-1)$
• D. $\frac{\sqrt{2}+1}{2}$

Answer 1

Answer: (C)

$2(\sqrt{2}-1)$

$I={\int }_0^{\pi /2}\frac{sinx}{\sqrt{1+cosx}}dx$

Let $t=1+cosx\Rightarrow dt=-sinxdx$

limit of t : when $x=0,t=2$

when $x=\frac{\pi }{2},t=1$

$\therefore I={\int }_2^1\frac{-dt}{\sqrt{t}}={\int }_1^2\frac{dt}{\sqrt{t}}$

$=2[\sqrt{t}{]}_1^2$

$=2[\sqrt{2}-1]$