Evaluate the integral

Question: 

$\oint_{0}^{a}9x^2\sqrt{a^2-x^2}$

Solution:

Assume $x = asin(t)$

            dx= a cos(t)dt

Limit would be,      $0 \le  t \le \pi/2$

Placing these values,

$\oint_{0}^{\pi/2}9(asin(t))^2\sqrt{a^2-((asin(t))^2}acos(t)dt$

$= \oint_{0}^{\pi/2}9(a^2sin^2(t))\sqrt{a^2-a^2sin^2(t)}acos(t)dt$

$= \oint_{0}^{\pi/2}9(a^2sin^2(t))a\sqrt{1-sin^2(t)}acos(t)dt$

$= 9a^4 \oint_{0}^{\pi/2}sin^2(t)\sqrt{cos^2(t)}cos(t)dt$                    $[sin^2(t) + cos^2(t) = 1]$

$= 9a^4 \oint_{0}^{\pi/2}sin^2(t)cos^2(t)dt$                            

$= 9a^4 \oint_{0}^{\pi/2} \frac{sin^2(2t)}{4}dt$                   $[sin(t)cos(t)= \frac{sin(2t)}{2}]$     

$= 9a^4 \oint_{0}^{\pi/2} \frac{1- cos(4t)}{8}dt$                $[sin^2(t) = \frac{1- cos(2t)}{2}]$

$= \frac{9a^4}{8} [(\pi/2 - \frac{sin(4\pi/2)}{4})-(0 - \frac{sin(0)}{4})]$       $ [\int{1-cos4t} dt = t - \frac{sin4t}{4}]$

$= \frac{9a^4}{8}(\frac{\pi}{2} - \frac{sin(2\pi)}{4})$

$= \frac{9a^4}{8}(\frac{\pi}{2} - 0)$

$=  \frac{9\pi a^4}{16}$