MCQ
$\int_{\, - 1/2}^{\,1/2} {(\cos x)\,\left[ {\log \left( {\frac{{1 - x}}{{1 + x}}} \right)} \right]\,dx = } $
- ✓$0$
- B$1$
- C${e^{1/2}}$
- D$2{e^{1/2}}$
$\therefore$ $I = \int_{ - 1/2}^{1/2} {\cos ( - x)\left[ {\log \left( {\frac{{1 + x}}{{1 - x}}} \right)} \right]} \,dx$
==> $I = - \int_{ - 1/2}^{1/2} {\cos x\left[ {\log \left( {\frac{{1 - x}}{{1 + x}}} \right)} \right]} \,dx$ ....$(ii)$
Adding $(i)$ and $(ii),$ we get
$2I = \int_{\, - 1/2}^{\,1/2} {\cos \,x\,\left[ {\log \,\left( {\frac{{1 - x}}{{1 + x}}} \right)} \right]} \,dx - \int_{\, - 1/2}^{\,1/2} {\cos \,x\,\left[ {\log \,\left( {\frac{{1 - x}}{{1 + x}}} \right)} \right]\,\,dx} $
or $2I = 0$ or $I = 0.$
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