Question

lf $f\left(x\right)=p\left(x\right)q\left(x\right)$ where $p\left(x\right)=\sqrt{\cos x},q\left(x\right)=\log \left(\frac{1-x}{1+x}\right)$ then ${\int }_a^{b}f\left(x\right)dx$ equals where $a=-1/2b=1/2$

• A. $2{\int }_0^{1/2}p\left(x\right)q\left(x\right)dx$
• B. ${\int }_0^{b}p\left(x\right)q\left(x\right)$
• C. 1
• D. 0

Answer 1

The correct option is D 0

$f\left(x\right)=p\left(x\right)q\left(x\right)$
$p\left(x\right)=\sqrt{\cos x},q\left(x\right)=\log \left(\frac{1-x}{1+x}\right)$
$I={\int }_{-1/2}^{1/2}\sqrt{\cos x}\log \left(\frac{1-x}{1+x}\right)dx$
$I={\int }_{-1/2}^{1/2}\sqrt{\cos (-x)}\log \left(\frac{1+x}{1-x}\right)dx$
$I=-{\int }_{-1/2}^{1/2}-\sqrt{\cos x}\log \left(\frac{1-x}{1+x}\right)dx$
$\Rightarrow I=-{\int }_{-1/2}^{1/2}\sqrt{\cos x}\log \left(\frac{1-x}{1+x}\right)dx$
$2I=0$
$I=0$