Question

If $I=\frac{\cot x}{\sqrt{a+b{\cot }^{2}x}}dx(0<a<b),$ then I equals

• A. $\frac{1}{\sqrt{b-a}}{\sin }^{-1}\left(\sqrt{\frac{b-a}{b}\sin x}\right)+Const$
• B. $\sqrt{b-a}{\sin }^{-1}\left(\sqrt{b-a}x\right)+Const$
• C. $\sqrt{b-a}{\sin }^{-1}\left(\sqrt{b-a}\sin x\right)+Const$
• D. $-\sqrt{b-a}{\cos }^{-1}\left(\sqrt{\frac{b-a}{b}\sin x}\right)+Const$

Answer 1

The correct option is A $\frac{1}{\sqrt{b-a}}{\sin }^{-1}\left(\sqrt{\frac{b-a}{b}\sin x}\right)+Const$

$I=\int \frac{\cot x}{\sqrt{a+b{\cot }^{2}x}}dx$

We know that, $\cot x=\frac{1}{\cos x}$

$I=\int \frac{\cos x}{\sqrt{a{\sin }^{2}x+b{\cos }^{2}x}}dx$

Also, ${\cos }^{2}x=1-{\sin }^{2}x$
Put $\sin x=t$
$I=\int \frac{dt}{\sqrt{b-(b-a)t^2}}=\frac{1}{\sqrt{b-a}}{\sin }^{-1}\left(\sqrt{\frac{b-a}{b}}t\right)+c$
$I=\frac{1}{\sqrt{b-a}}{\sin }^{-1}\left(\sqrt{\frac{b-a}{b}}\sin x\right)+c$