Evaluate: `int(cos^3x)/(sin^2x+sinx)dx`

Explanation

Given,
`I=int(cos^3x)/(sin^2x+sinx)dx`
`I=int((cos^2x)cos x)/(sin^2x+sinx) dx`
`=int((1-sin^2x)cos x)/(sin^2x+sinx) dx`
Put `t=sinx `

`=>cosxdx=dt`
so,`I=int(1/t-1)dt`
`=logt−t+C`

`=>I=logsinx−sinx+C`

`=>I=logsinx−sinx ` (Given` C=0`)
Therefore, `f(pi/2)=-1`