How do you integrate $int x^3 sec^2 x dx$ using integration by parts?

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Answer 1 · 2016-03-14T21:47:07

The first two steps are OK. After that you need polylogarithmic functions (beyond the scope of an introductory course).

Let $I = int x^3 sec^2 x dx$

Let $u=x^3$ and $dv = sec^2x dx$ .

This makes $du = 3x^2$ and $v = int sec^2x dx = tanx$

So, we have

$uv-int v du = x^3 tanx - 3 int x^2 tanx dx$

To evaluate $int x^2 tanx dx$ use parts again.

Let $u = x^2$ and $dv = tanx dx$ .

So that $du = 2x dx$ and $v = int tanx dx = -ln abs cosx$

So we have

$I = x^3 tanx -3[-x^2 ln abs cosx+2 int x ln abs cosx dx]$

The last integral involves the polylogarithmic function.

I don't know enough about that to explain it. (Good luck.)

How do you integrate int x^3 sec^2 x dx int x^3 sec^2 x dx using integration by parts?