Question

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

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

Answer 1

Answer: (C)

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