MCQ
$\int_0^{\pi /2} {\frac{{{{\sin }^{3/2}}x\,dx}}{{{{\cos }^{3/2}}x + {{\sin }^{3/2}}x}}} = $
  • A
    $0$
  • B
    $\pi $
  • C
    $\pi /2$
  • $\pi /4$

Answer

Correct option: D.
$\pi /4$
d
(d) Let $I = \int_0^{\pi /2} {\frac{{{{\sin }^{3/2}}x\,dx}}{{{{\cos }^{3/2}}x + {{\sin }^{3/2}}x}}} $.....$(i)$

$=  \int_0^{\pi /2} {\frac{{{{\sin }^{3/2}}\left( {\frac{\pi }{2} - x} \right)}}{{{{\cos }^{3/2}}\left( {\frac{\pi }{2} - x} \right) + {{\sin }^{3/2}}\left( {\frac{\pi }{2} - x} \right)}}dx} $

$=  \int_0^{\pi /2} {\frac{{{{\cos }^{3/2}}x\,dx}}{{{{\sin }^{3/2}}x + {{\cos }^{3/2}}x}}} $.....$(ii)$

Adding $(i)$ and $(ii),$ we get $I = \frac{1}{2}\int_0^{\pi /2} {1dx = \frac{1}{2}[x]_0^{\pi /2} = \frac{\pi }{4}} $.

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