Question

how to integrate this

Answer 1

\[\int\limits \sin ^{\frac {2}{3}}x \cos^3x dx\]

Answer 2

Let u = sinx , du = cos x dx
moreover, cos^2x = 1 - sin^2 x
so that you have
\[\int u^{2/3}(1-u^2) du\]
quite easy now, right?

Answer 3

yes, thanks @OOOPS

Answer 4

:)