`int x tan^2(x) dx` Evaluate the integral

`intxtan^2(x)dx`

Let's evaluate the above integral by using the method of integration by parts,

If f(x) and g(x) are differentiable functions , then

`intf(x)g'(x)=f(x)g(x)-intf'(x)g(x)dx`

If we write f(x)=u and g'(x)=v, then

`intuvdx=uintvdx-int(u'intvdx)dx`

Let's denote u=x and `v=tan^2(x)`

`intxtan^2(x)dx=x*inttan^2(x)dx-int(1inttan^2(x)dx)dx`

Now let's use the identity:`tan^2(x)=sec^2(x)-1`

`=x*int(sec^2(x)-1)dx-int(int(sec^2(x)-1)dx)dx`

`=x*(intsec^2(x)dx-int1dx)-int(int(sec^2(x)-1)dx)dx`

`=x*(tan(x)-x)-int(tan(x)-x)dx`

`=xtan(x)-x^2-inttan(x)+intxdx`

`=xtan(x)-x^2-(-ln|cos(x)|)+x^2/2`

`=xtan(x)-x^2+x^2/2+ln|cos(x)|`

`=xtan(x)+ln|cos(x)|-x^2/2`

Add a constant C to the solution,

`=xtan(x)+ln|cos(x)|-x^2/2+C`

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