Answer
$-\frac{4x^3-3sin(2x)(2x^2-1)-6xcos(2x)}{24}+C$
Work Step by Step
$\int x^2 (sinx)^2dx= \int x^2 (\frac{1-cos(2x))}{2x}dx $
$=\frac{1}{2} \int x^2dx$ $-\frac{1}{2}\int x^2 cos(2x)dx$
for $\int x^2 cos(2x)dx$,
$let u=x^2$
$du=2xdx$
$dv=cos(2x)dx$
$v=\frac{1}{2}sin(2x)$
apply integration by parts
$\int udv=uv-\int vdu$
$=\frac{x^2 (sin2x)}{2}- \int \frac{1}{2}sin(2x)2xdx$
$=\frac{x^2 (sin2x)}{2}- \int x sin(2x)dx$
for $\int x sin(2x)dx,$
$let u=x$
$du=dx$
$dv=sin(2x)dx$
$v=-\frac{cos(2x)}{2}$
apply integration by parts
$\int udv=uv-\int vdu$
$=-\frac{xcos(2x)}{2}- \int -\frac{1}{2} cos(2x)dx$
$=-\frac{xcos(2x)}{2}+\frac{1}{2} \int cos(2x)dx$
$=-\frac{xcos(2x)}{2}+\frac{sin(2x)}{4}$
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$for \frac{1}{2} \int x^2dx,$
$\frac{1}{2} \int x^2dx= \frac{x^3}{6}$
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$\frac{1}{2} \int x^2dx$ $-\frac{1}{2}\int x^2 cos(2x)dx$
$=\frac{x^3}{6}-\frac{1}{2}(\frac{sin(2x)(2x^2-1)+2xcos(2x)}{4})+C$
$=-\frac{4x^3-3sin(2x)(2x^2-1)-6xcos(2x)}{24}+C$
