Question

Evaluate: ${\int }_0^1\cos$ $\left(2{\cot }^{-1}\sqrt{\frac{1-x}{1+x}}\right)dx$

• A. $\frac{-1}{2}$
• B. $\frac{1}{2}$
• C. $0$
• D. $1$

Answer 1

The correct option is A $\frac{-1}{2}$

Let, $x=\cos \theta ⟹dx=-sin\theta d\theta$

And, $\theta ={\cos }^{-1}x$

At, $x=0,\theta =\frac{\pi }{2}$ and at $x=1,\theta =0$

Now, $\sqrt{\frac{1-x}{1+x}}=\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}$

$=\sqrt{\frac{2{\sin }^2\frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}}$

$=\tan \frac{\theta }{2}=\cot (\frac{\pi }{2}-\frac{\theta }{2})$

$⟹2{\cot }^{-1}\sqrt{\frac{1-x}{1+x}}=(\pi -\theta )$

Hence, integration becomes:-

${\int }_{\pi /2}^0\cos (\pi -\theta )(-\sin \theta )d\theta$

$={\int }_{\pi /2}^0\sin \theta \cos \theta d\theta$

$=\frac{1}{2}{\int }_{\pi /2}^0\sin 2\theta .d\theta$

$=\frac{-1}{4}{\left[\cos 2\theta \right]}_{\pi /2}^0$

$=\frac{-1}{4}(1+1)=\frac{-1}{2}$

Answer-(A)