Question

If $\int \frac{2dx}{\sqrt{{\cot }^{2}x-{\tan }^{2}x}}=-\sqrt{f(x)}+c$, then $f(x)=$

• A. $\cot x$
• B. $\sin 2x$
• C. $\cos 2x$
• D. $tanx$

Answer 1

Answer: (C)

We have,
$\int \frac{2dx}{\sqrt{{\cot }^{2}x-{\tan }^{2}x}}dx=\int \frac{2dx}{\sqrt{\frac{{\cos }^{2}x}{{\sin }^{2}x}-\frac{{\sin }^{2}x}{{\cos }^{2}x}}}$
$=\int \frac{2\sin x\cos x}{\sqrt{{\cos }^{4}x-{\sin }^{4}x}}dx[\because \sin 2x=2\sin x\cos x]$
$=\int \frac{\sin 2xdx}{\sqrt{\left({\cos }^{2}x-{\sin }^{2}x\right)\left({\cos }^{2}x+{\sin }^{2}x\right)}}$
$\left[\because {\sin }^{2}x+{\cos }^{2}x=1\text{ and }{\cos }^{2}x-{\sin }^{2}x=\cos 2x\right]$
$=\int \frac{\sin 2xdx}{\sqrt{\cos 2x}}$
Put $\cos 2x=t$
$\Rightarrow -2\sin 2xdx=dt$
$\Rightarrow -\frac{1}{2}\int \frac{dt}{\sqrt{t}}=-\frac{1}{2}\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}+c$
$$\begin{gathered} \Rightarrow -\frac{1}{2}\frac{t^{1/2}}{\frac{1}{2}}+c=-\sqrt{t}+c \\ =-\sqrt{\cos 2x}+c \end{gathered}$$
$[t=\cos 2x]$
Compare with $-\sqrt{f(x)}+c$
So, $f(x)=\cos 2x$