e^x ( 1 - x 1 - x ) , ,dx is equal to - Vidyadip

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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