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?

$\int (secxtanx)(tanx)~dx$

$= secx(tanx) - \int secx(sec^{2})~ dx$

$= secxtanx-\int sec^{3}x~dx$

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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