MCQ
If $\int_{}^{} {\frac{{{e^x}(1 + \sin x)dx}}{{1 + \cos x}} = {e^x}f(x) + c} $, then $f(x) = $
  • A
    $\sin \frac{x}{2}$
  • B
    $\cos \frac{x}{2}$
  • $\tan \frac{x}{2}$
  • D
    $\log \frac{x}{2}$

Answer

Correct option: C.
$\tan \frac{x}{2}$
c
$I = \int_{}^{} {{e^x}\left( {\frac{{1 + \sin x}}{{1 + \cos x}}} \right)\,dx}  = \int_{}^{} {{e^x}\left[ {\frac{{1 + 2\sin (x/2)\,\cos (x/2)}}{{2{{\cos }^2}(x/2)}}} \right]dx} $

$I = \int_{}^{} {{e^x}\left[ {\frac{1}{2}{{\sec }^2}(x/2) + \tan (x/2)} \right]\,dx}  = {e^x}.\tan (x/2) + c$

$\{ \,\,\,\int_{}^{} {{e^x}[f(x) + f'(x)\,]dx = {e^x}.\,f(x) + c\} } $

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