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:

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

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