# $I= \int \limits_0^1 \int x^5 (1-x^2)^{\frac{3}{2}}dx$ Find the value of I

$\begin {array} {1 1} (a)\;\frac{11}{315} \\ (b)\;\frac{8}{315} \\ (c)\;\frac{1}{720} \\ (d)\;\frac{8}{945} \end {array}$

$x= \sin t$
$dx= \cos t dt$
$\int \limits_0^{\frac{\pi}{2}} \sin^5 \cos^ 4 t dt$
$\cos t=k$
$-\sin t dt =dk$
$\qquad= -\int \limits_1^0 k^4 (1-k^2)^2 dk$
$\qquad= -\int \limits_1^0 k^4 (k^4+1-2k^2) dk$
$\qquad= -\int \limits_1^0 (k^8+k^4-2k^6) dk$
$\qquad= \large\frac{1}{9}+\frac{1}{5} -\frac{2}{7}$
$\qquad= \large\frac{35+63-90}{315}$
$\qquad= \large\frac{8}{315}$
Hence b is the correct answer.