We shall now proceed to integrating secx and similarly, lets refresh the formulas we should know.

\frac {d}{dx}tanx = sec^{2}x

\frac {d}{dx}secx = secxtanx

\int tanx dx = ln|secx|+c (MF15)

\int secx dx = ln|secx+tanx|+c (MF15)

\int sec^{2}x dx = tanx+c

\int sec^{3}x dx = \int secx(sec^{2}x)dx = \int secx(tan^{2}x+1)dx = \int secxtan^{2}x+secx dx
So how do we \int secxtan^{2}x dx? I’ll first rewrite it as \int (secxtanx)(tanx)dx for some insights.

We can’t adopt the \int f'(x)f(x) dx method here. So, Integration by parts?

\int (secxtanx)(tanx)dx = secx(tanx) - \int secx(sec^{2}) dx = secxtanx-\int sec^{3}xdx

Wait! \int sec^{3}x dx again? hmmm.

so we have that \int sec^{3}x dx = \int secxtan^{2}x+secx dx = secxtanx-\int sec^{3}xdx + \int secx dx. Then with a bit of juggling and manipulations, we have
2\int sec^{3}x dx = secxtanx + ln|secx+tanx|+c.

I do hope this gives you some insights. You should try \int sec^{4}x dx on your own using the information here.

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