STD 12 - 7.2 definite integral — JEE कक्षा 11 साइन्स — Question

$\Rightarrow dx = 2a\sin 2\theta \,d\theta $रखने पर

$\therefore$  $\int_0^a {x{{(2ax - {x^2})}^{3/2}}dx} $

$ = \int_0^{\pi /4} {2{a^5}(1 - \cos 2\theta ){{\sin }^4}2\theta \,\,d\theta } $

पुन: $2\theta = \phi $ रखने पर

$ = {a^5}\left[ {\int_0^{\pi /2} {{{\sin }^4}\phi \,d\phi } - \int_0^{\pi /2} {{{\sin }^4}\phi \cos \phi \,d\phi } } \right]$

$ = {a^5}\left[ {\frac{{3\pi }}{{16}} - \frac{1}{5}} \right]$.