Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 49

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