Question

If $\int \sqrt{1+\text{cosec}x}dx=f\left(x\right)+C$, where $C$ is a constant of integration, then $f\left(x\right)$ is equal to

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

Answer 1

The correct option is A ${\cos }^{-1}(1-2\sin x)$

$I=\int \sqrt{1+\text{cosec}x}dx$
$=\int \sqrt{\frac{\sin x+1}{\sin x}}dx$
$=\int \sqrt{\frac{(1+\sin x)(1-\sin x)}{\sin x(1-\sin x)}}dx$
$=\int \frac{\cos x}{\sqrt{\frac{1}{4}-{(\frac{1}{2}-\sin x)}^2}}dx$
Put $\frac{1}{2}-\sin x=t$
Then $I=\int \frac{-dt}{\sqrt{{\left(\frac{1}{2}\right)}^2-t^2}}$
$=-{\sin }^{-1}\left(\frac{t}{1/2}\right)+c$
$=-{\sin }^{-1}(1-2\sin x)+c$
$={\cos }^{-1}(1-2\sin x)-\frac{\pi }{2}+c$
$={\cos }^{-1}(1-2\sin x)+C$ where $C=c-\frac{\pi }{2}$