MCQ
$\int_{\pi /4}^{\pi /2} {\cos \theta \,{\rm{cose}}{{\rm{c}}^{\rm{2}}}\theta \,d\theta = } $
  • $\sqrt 2 - 1$
  • B
    $1 - \sqrt 2 $
  • C
    $\sqrt 2 + 1$
  • D
    None of these

Answer

Correct option: A.
$\sqrt 2 - 1$
a
(a) Let $\int_{\pi /4}^{\pi /2} {\cos \theta \frac{1}{{{{\sin }^2}\theta }}d\theta } $

Put $t = \sin \theta \Rightarrow dt = \cos \theta \,\,d\theta ,$

then we have

$\int_{1/\sqrt 2 }^1 {\frac{1}{{{t^2}}}dt} = \left[ {\frac{{ - 1}}{t}} \right]_{1/\sqrt 2 }^1 $

$= \sqrt 2 - 1$.

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