Question

The value of integral $\int \sqrt{1+\text{cosec}x}dx$ is equal to

Note: $C$ is the constant of integration.

• A. $2{\sin }^{-1}\sqrt{\sin x}+C$
• B. $2{\cos }^{-1}\sqrt{\cos x}+C$
• C. $C-2{\sin }^{-1}(1-2\sin x)$
• D. ${\cos }^{-1}(1-2\sin x)+C$

Answer 1

Answer: (A), (D)

$2{\sin }^{-1}\sqrt{\sin x}+C$
${\cos }^{-1}(1-2\sin x)+C$

$\int \sqrt{1+cosecx}dx$
$I=\int \frac{\sqrt{(1+\sin x)(1-\sin x)}}{\sqrt{\sin x(1-\sin x)}}dx$
$=\int \frac{\cos x}{\sqrt{\sin x(1-\sin x)}}dx$
$=\int \frac{\cos x}{\sqrt{\frac{1}{4}-(\frac{1}{2}-\sin x{)}^2}}dx$
Putting $\frac{1}{2}-\sin x=t$
$\Rightarrow -\cos xdx=dt$
$=\int \frac{-dt}{\sqrt{{\left(\frac{1}{2}\right)}^2-t^2}}$
$=-{\sin }^{-1}\left(\frac{t}{1/2}\right)+C=-{\sin }^{-1}(1-2\sin x)+C$
$={\cos }^{-1}(1-2\sin x)+C-\frac{\pi }{2}$
$={\cos }^{-1}(1-2\sin x)+C$
$={\cos }^{-1}(1-2(\sqrt{\sin x}{)}^2)+C$
$={\cos }^{-1}(1-2{\sin }^{2}t)+C$
Putting $\sqrt{\sin x}=\sin t$
$={\cos }^{-1}\left(\cos 2t\right)+C$............ $[\because \sqrt{\sin x}>0\Rightarrow tϵ(0,\frac{\pi }{2})]$
$=2t+C$
$=2{\sin }^{-1}\left(\sqrt{\sin x}\right)+C$