Question

The value of the integral $Ι=\displaystyle \int \frac{d x}{\sqrt{1 + s i n x}}, \, \forall x\in \left[0 , \frac{\pi }{2}\right]$ 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 $=\int \frac{d x}{\sin \frac{x}{2}+\cos \frac{x}{2}}$
$=\frac{1}{\sqrt{2}} \int \frac{d x}{\sin \left(\frac{x}{4}+\frac{x}{2}\right)}$
$=\frac{1}{\sqrt{2}} \int \operatorname{cosec}\left(\frac{\pi}{4}+\frac{x}{2}\right) d x$
$=\frac{1}{\sqrt{2}} 2 \ln \left(\tan \left(\frac{\pi}{8}+\frac{x}{4}\right)\right)+c$
$=\sqrt{2} \ln \left(\tan \left(\frac{\pi}{8}+\frac{x}{4}\right)\right)+c$
$\Rightarrow k=\sqrt{2}$
Hence, $\sqrt{2} k=2$