Question :
$\displaystyle \int sin^4.cos^2x.dx$
Attempt :
I reached a step where >>
$\displaystyle \frac{1}{8} \int (1-2cos2x+cos^2x).(1+cos2x)$
How to expand this ?
$\displaystyle \int{sin^4xcos^2x}dx=\int{sin^2xsin^2xcos^2x}dx$
$\displaystyle =\int{\left(1-cos^2x\right)\left(1-cos^2x\right)cos^2x}dx$
$\displaystyle =\int{\left(1-2cos^2x+cos^4x\right)cos^2x}dx$
may be more straightforward.
Also you have erred on the way...
$\displaystyle sin^2xsin^2x=\frac{1}{4}(1-cos2x)(1-cos2x)=\frac{1}{4}\left(1-2cos2x+cos^2(2x)\right)$