We have,
∫π20sin2xcosxdx
∫π202sinxcosxcosxdx
⇒2∫π20sinxcos2xdx
Let
cosx=t
sinxdx=dt
Therefore,
2∫01t2dt
On integrating and we get,
=2[t33]10
Then,
=2[03−133]
=−23
Hence, this is the answer.