Question

Evaluate integral `xtan^2xdx`?

Answer 1

`I=xtanx-ln|secx|-x^2/2+C`

Explanation:

`int(u*v)dx=u*intvdx-int (u^'intvdx)dx...to(1)`
Here,
`I=intxtan^2xdx=intx(sec^2x-1)dx`

`=intxsec^2xdx-intxdx,`

Using Integration by parts (1) above.
`u=xandv=sec^2x=>u^'=1andintvdx=intsec^2xdx=tanx`

`I=[x*intsec^2xdx-int(d/(dx)(x)intsec^2xdx)dx]-x^2/2+c_1`

`=>I=[x*tanx-int(1)tanxdx]-x^2/2+c_1`
`=>I=[xtanx-ln|secx|+c_2]-x^2/2+c_1`
`I=xtanx-ln|secx|-x^2/2+C,where,C=c_1+c_2`