Integrating Trigonometric functions (part 3)

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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