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.)