Question

$\int \frac{dx}{sin2x\sqrt{{cot}^{2}x-1}}$ equals (when $0$ < $x$ < $\frac{\pi }{4}$)

• A. $2{sec}^{-1}$ $\left(cotx\right)$ $+$ $c$
• B. $\frac{1}{2}{sec}^{-1}\left(cotx\right)+c$
• C. $\frac{1}{2}{cosec}^{-1}\left(cotx\right)+c$
• D. $None$ $of$ $these$

Answer 1

The correct option is C $\frac{1}{2}{cosec}^{-1}\left(cotx\right)+c$

$\int \frac{dx}{\sin 2x\sqrt{{\cot }^{2}x-1}}$

$=\int \frac{dx}{2\sin x\cos c\sqrt{{\cot }^{2}x-1}}$

dividing by ${\sin }^{2}x$ to both Nr. and Dn.

$\Rightarrow =\frac{1}{2}\int \frac{cose{c}^{2}xdx}{\cot x\sqrt{{\cot }^{2}x-1}}$

Let $\cot x=t$

$\Rightarrow \frac{dt}{dx}=-cose{c}^{2}x$

$\Rightarrow cose{c}^{2}xdx=-dt$

$\Rightarrow =\frac{1}{2}\int \frac{-dt}{t\sqrt{t^2-1}}$

$=\frac{1}{2}cose{c}^{-1}\left(t\right)+c$

$=\frac{1}{2}cose{c}^{-1}\left(\cot x\right)+c$.