How do you integrate #int x sec 5 x dx # using integration by parts?

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Answer 1 · 2016-02-26T14:05:42

I don't believe you do.

We can start with #u=x# and #dv=sec(5x)#, so that

#du = dx# and #v = int sec(5x) dx = 1/5 ln(tan5x+sec5x)#

We get,

#int x sec(5x) dx = x/5 ln(tan5x+sec5x) - int ln(tan5x+sec5x) dx#

If you have some way of finding that integral, use it.

Wolfram Alpha gives the original integral as:

#int x sec(5 x) dx = 1/25 i (Li_2(-i e^(5 i x))-Li_2(i e^(5 i x)))+1/5 x (log(1-i e^(5 i x))-log(1+i e^(5 i x)))+C#

Where #Li_n# is the polylogarithmic function.

You can read more about the polylogarithmic function here: