Question

If $\int \sqrt{\frac{1-x}{1+x}}dx=\sqrt{1-x^2}+f\left(x\right)+c,x\in [0,1)$ where $f\left(0\right)=-\frac{\pi }{2}$ then $f\left(\frac{1}{2}\right)$ is ______

• A. $-\frac{\pi }{3}$
• B. $\frac{\pi }{3}$
• C. $-\frac{\pi }{6}$
• D. None of these

Answer 1

The correct option is A $-\frac{\pi }{3}$

Solution -

$=\int \sqrt{\frac{1-x}{1+x}}\times \frac{\sqrt{1-x}}{\sqrt{1-x}}dx$

$=\int \frac{1-x}{\sqrt{1-x^2}}dx=\int \frac{dx}{\sqrt{1-x^2}}-\int \frac{xdx}{\sqrt{1-x^2}}$

$=-co{s}^{-1}x-\int \frac{xdx}{\sqrt{1-x^2}}$

put $1-x^2=t$

$-2xdx=dt$

$=-co{s}^{-1}x+\frac{1}{2}\int \frac{dt}{\sqrt{t}}$

$=-co{s}^{-1}x+\sqrt{1-x^2}+c$

$=f\left(x\right)=-co{s}^{-1}x$

$f\left(0\right)=\frac{-\pi }{2}$ $f\left(\frac{1}{2}\right)=-co{s}^{-1}\left(\frac{1}{2}\right)=-\frac{\pi }{3}$

A is correct