Question

The value of the integral $I=\int \frac{dx}{\sqrt{1+sinx}},\forall x\in \left[0,\frac{\pi }{2}\right]$ is equal to $k\ln \left(\tan \left(\frac{\pi }{4}+\frac{x}{8}\right)\right)+c,$ then the value of $k\sqrt{2}$ is equal to (where, $c$ is the constant of integration)

• A. $\sqrt{2}$
• B. $\frac{1}{2}$
• C. $2$
• D. $2\sqrt{2}$

Answer 1

Answer: (C)

As, $1+\sin x={\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}$,
$={\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)}^2$
I $=\int \frac{dx}{\sin \frac{x}{2}+\cos \frac{x}{2}}$
$=\frac{1}{\sqrt{2}}\int \frac{dx}{\sin \left(\frac{x}{4}+\frac{x}{2}\right)}$
$=\frac{1}{\sqrt{2}}\int \mathrm{cosec}\left(\frac{\pi }{4}+\frac{x}{2}\right)dx$
$=\frac{1}{\sqrt{2}}2\ln \left(\tan \left(\frac{\pi }{8}+\frac{x}{4}\right)\right)+c$
$=\sqrt{2}\ln \left(\tan \left(\frac{\pi }{8}+\frac{x}{4}\right)\right)+c$
$\Rightarrow k=\sqrt{2}$
Hence, $\sqrt{2}k=2$