MCQ
$\int\limits_0^\pi  {\,\,\frac{{x\cos x}}{{{{\left( {1 + \sin x} \right)}^2}}}} dx$ is equal to :
  • A
    $\pi -2$
  • B
    $- (2 + \pi )$
  • C
    zero
  • $2 -\pi$

Answer

Correct option: D.
$2 -\pi$
d
Take $x$ as the first and $\frac{{\cos \,\,x}}{{{{\left( {1\,\, + \,\,\sin \,x} \right)}^2}}}$ as the second function. $I B P$

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