Question

integrate tan^8 (2x)

Answer 1

\[\Large \int \tan^n(u) du = \frac{1}{n-1}\tan^{n-1}(u)-\int \tan^{n-2}(u) du\]

Answer 2

Might wanna do a u-sub on 2x

Answer 3

what method is that i cant recognize...

Answer 4

It's just a general integral for tan^n that comes in handy when u have these bigger ones

\[\int \tan^8(2x) dx = \frac{1}{2} \int \tan^8(u) du \mbox{ where } u = 2x\]
\[\frac{1}{2}\left(\frac{1}{7}\tan^7(u)-\int \tan^6(u) du\right)\]
You can keep reducing the tan^m until you're left with tan^2 which can be solved easily
\[\frac{1}{2}\left(\frac{1}{7}\tan^7(u) - \frac{1}{5}\tan^5(u) + \int \tan^4(u) du\right)\]
\[\frac{1}{2}\left(\frac{1}{7}\tan^7(u) - \frac{1}{5}\tan^5(u) + \frac{1}{3}\tan^3(u)-\int \tan^2(u) du\right)\]
\[\frac{1}{2}\left(\frac{1}{7}\tan^7(u) - \frac{1}{5}\tan^5(u) + \frac{1}{3}\tan^3(u)-\tan(u) + u\right) = \]
\[\frac{\tan^7(2x)}{14} - \frac{\tan^5(2x)}{10}+\frac{\tan^3(2x)}{6}-\frac{\tan(2x)}{2}+\frac{2x}{2} + c\]


I'm pretty sure there are other approaches but this one would be the easiest to do.

If you want me to explain why the formula works just ask :)

Answer 5

oh.... i see...
may you teach me integrating that using trig. solvable groups of trig. integrals?

Answer 6

Basically using that tan^2(u) = sec^2(u) - 1:
\[ \int \tan^8(2x) dx = \int \tan^6(2x)(\sec^2(2x) - 1) dx = \int \tan^6(2x)\sec^2(2x) dx - \int tan^6(2x) dx\]
The integral tan^6(2x)sec^2(2x) dx can be done using u-sub tan^(2x), du/2 = sec^2(2x) dx

for the \(\int \tan^6(2x) dx\), the process is repeated by stripping out tan^2(2x) again to turn it into \(\int \tan^4(2x)(\sec^2(2x) -1)dx = \int \tan^4(2x)\sec^2(2x)dx - \int \tan^4(2x)dx\)

And again on the tan^4(2x), so the integral becomes:
\(\int \tan^6(2x)\sec^2(2x) dx - \int \tan^4(2x)\sec^2(2x)dx + \int \tan^2(2x)\sec^2(2x)dx - \int \tan^2(2x)dx\)
Each integral can be done relatively easy using u-sub
\[\frac{\tan^7(2x)}{14}-\frac{\tan^5(2x)}{10} + \frac{\tan^3(2x)}{6}-\frac{\tan(2x)}{2}+\frac{2x}{2}+c\]

Answer 7

different approach not using formula^, the formula is derived from this though

Answer 8

thanks genius! :)