STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

$\Rightarrow dx = a\cos \theta \,\,d\theta $

Now $\int_0^a {{x^4}\sqrt {{a^2} - {x^2}} } dx = {a^6}\int_0^{\pi /2} {{{\sin }^4}\theta \cos \theta \cos \theta \,d\theta } $

$ = {a^6}\int_0^{\pi /2} {{{\sin }^4}\theta\, {{\cos }^2}\theta \,d\theta } $

$ = {a^6}\frac{{\Gamma \left( {\frac{5}{2}} \right).\Gamma \left( {\frac{3}{2}} \right)}}{{2\Gamma 4}} = \frac{\pi }{{32}}{a^6}$,

(Using gamma function).