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7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
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
    $2{e^{1/2}}$
✓

Answer

Correct option: A.
$0$
a
(a) $I = \int_{ - 1/2}^{1/2} {(\cos x)\left[ {\log \left( {\frac{{1 - x}}{{1 + x}}} \right)} \right]dx} $ ....$(i)$

$\therefore$  $I = \int_{ - 1/2}^{1/2} {\cos ( - x)\left[ {\log \left( {\frac{{1 + x}}{{1 - x}}} \right)} \right]} \,dx$

==> $I = - \int_{ - 1/2}^{1/2} {\cos x\left[ {\log \left( {\frac{{1 - x}}{{1 + x}}} \right)} \right]} \,dx$ ....$(ii)$

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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