So part 3, we look at $tanx$! This is slightly more complicated as we need to first lay out some formulas that 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)

Notice how $tanx$ and $secx$ are quite related here… And we recall we have a trigonometry identity that happens to be that closely related:

$tan^{2}x+1=sec^{2}x$!

$\int tan^{2}x dx = \int sec^{2}x -1 ~dx = tanx - x + c$

$\int tan^{3}x ~dx$

$= \int tanx(tan^{2}x) ~dx$

$= \int tanx(sec^{2}x-1) ~dx$

$= \int tanxsec^{2}x-tanx ~dx$

Here we have a slight problem with $\int tanxsec^{2}x$

Recall the $\int f'(x)f(x)$ method we saw in part 1 and 2…
$\int tanxsec^{2}x= \frac {tan^{2}x}{2}+c$

So, $\int tanxsec^{2}x-tanx ~dx=\frac {tan^{2}x}{2}-ln|secx|+c$

$\int tan^{4}x dx = \int tan^{2}x(sec^{2}x-1) ~dx = \int tan^{2}xsec^{2}x-tan^{2}x dx$

As you probably guessed, $\int tan^{2}xsec^{2}x ~dx = \frac {tan^{3}x}{3}+c$ and for $\int tan^{2}x dx$, we can refer to the previous results.

We will look at $secx$ in part 4, and surprise surprise, it will be quite a different appraoch! Let me know if there are any questions.

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