Question

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

• A. $-1/2$
• B. $1/2$
• C. $0$
• D. $1$

Answer 1

The correct option is C $0$

Let $\theta ={\cot }^{-1}\frac{\sqrt{1-x}}{\sqrt{1+x}}=\frac{B}{p}$

$\Rightarrow {\cos }^1\frac{B}{H}={\cos }^{-1}\frac{\sqrt{1-x}}{\sqrt{1-x+1+x}}$

$\Rightarrow \theta ={\cos }^{-1}\sqrt{1-x}$

$\Rightarrow {\int }_0^1\cos 2\theta dx$

$={\int }_0^{1}2{\cos }^2\theta -1dx$

$={\int }_0^{1}2{\cos }^2\left({\cos }^{-1}\sqrt{1-x}\right)-1dx$

$\Rightarrow {\int }_0^{1}2(1-x)-1dx$

$={\int }_0^{1}2-2x-1dx$

$={\int }_0^1-2x+1dx$

$=\frac{-2x^2}{2}+x{\int }_0^1$

$=-1+1=0$

$\Rightarrow \left(C\right)$.