$\displaystyle
\int sin^2(x) cos^3(x) dx
$
My answer came out to be?
$\displaystyle
sin^5x/5 - sin^3x/3
$
Write the integral as $\displaystyle \int sin^2(x) cos^2(x) cos(x)dx= \int sin^2(x)(1- sin^2(x)) cos(x)dx$$\displaystyle = \int (sin^2(x)- sin^4(x)) cos(x)dx$
Now let u= sin(x) so that du= cos(x)dx and the integral becomes $\displaystyle \int (u^2- u^4)du$$\displaystyle \frac{u^3}{3}- \frac{u^5}{5}+ C= \frac{sin^3(x)}{3}- \frac{sin^5(x)}{5}$, the negative of your answer.
It is always better, in asking for a check, to show your work.