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7.1 inderfinite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.1 inderfinite integral1 Mark
MCQ
$\int {{e^x}\left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)\,\,dx} $ is equal to
  • A
    $ - {e^x}\tan \,\left( {x/2} \right)$
  • ✓
    $ - {e^x}\cot \,\left( {x/2} \right)$
  • C
    $ - \frac{1}{2}{e^x}\tan \,\left( {\frac{x}{2}} \right)$
  • D
    $\frac{1}{2}{e^x}\cot \,\left( {\frac{x}{2}} \right)$
✓

Answer

Correct option: B.
$ - {e^x}\cot \,\left( {x/2} \right)$
b
(b) $I = \int {{e^x}\left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)dx} $$ = \int {{e^x}\left( {\frac{{1 - \sin x}}{{2{{\sin }^2}(x2)}}} \right)\,dx} $
==> $I = \int {{e^x}\left( {\frac{1}{2}{\rm{cose}}{{\rm{c}}^2}\frac{x}{2} - \cot \frac{x}{2}} \right)} \,dx$
$\left( \because \,\,\int{{{e}^{x}}\left( f(x)+f'(x) \right)}={{e}^{x}}f(x)+c \right)$
$\therefore \,\,I = {e^x}\left( { - \cot \frac{x}{2}} \right) + c = - {e^x}\cot \frac{x}{2} + c$.

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