Evaluate : `intxcos^(3)xsinxdx`.

Explanation

Take x as first function and `(cos^(3)xsinx)` as the second.
Putting cos x=t, we can evaluate `intcos^(3)xsinxdxas-(1)/(4)cos^(4)x`.
So, integrating by parts, we get
`intxcos^(3)x sindx=x*((-1)/(4)cos^(4)x)-int1*((-1)/(4))cos^(4)xdx`
`-(x)/(4)cos^(4)int((1+cos2x)/(2))^(2)dx`
`=-(xcos^(4)x)/(4)+(1)/(4)int((1)/(4)+(cos^(2)2x)/(4)+cos2x)dx`
`=-(xcos^(4)x)/(4)+(1)/(16)intdx+(1)/(4)2xdx+(1)/(32)int2cos^(2)2xdx`
`=-(xcos^(4)x)/(4)+(x)/(16)+(sin2x)/(8)+(1)/(32)int(1+cos4x)dx+C`
`=-(xcos^(4)x)/(4)+(x)/(16)+(sin2x)/(8)+(1)/(32)intcos4xdx`
`=-(xcos^(4)x)/(4)+(3x)/(32)+(sin2x)/(8)+(sin4x)/(128)+C`.