How do you evaluate the integral #int xtan^2x#?

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Answer 1 · 2017-03-09T07:12:12

I would begin by using the identity #tan^2(theta)=sec^2(theta)-1#.

#=>intx(sec^2(x)-1)dx#

#intxsec^2(x)dx-intxdx#

The RH is a basic integral. Let's continue with the left. Now we will use integration by parts.

#u=x color(white)(space) dv=sec^2(x)dx#

#du=dxcolor(white)(space)v=tan(x)#

Our integral now takes the form of #uv-intvdu#:

#xtan(x)-inttan(x)dx#

The RH is a basic integral.

Combining this with what we found above:

#xtan(x)-inttan(x)dx-intxdx#

Integrating, we get:

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