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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