Question

Little help please?

∫tan^6(2x)dx

Answer 1

expand and usub?

Answer 2

\[\int\limits \tan^6(2x) dx\] right

Answer 3

So substitute u = 2x, du = 2dx
\[\frac{ 1 }{ 2 } \int\limits \tan^6(u) du\] I think then you'll have to use a reduction formula

Answer 4

\[\int\limits \tan^nx dx = ((\tan^n-1)x)/(n-1) - \int\limits \tan ^{n-2} xdx\]

Answer 5

Yeah that's it

Answer 6

You'll have to use that twice, n = 6, and n = 4

Answer 7

And then you can write tan^u = sex^2u-1

Answer 8

sec*

Answer 9

tan^u being the second part being integrated?

Answer 10

\[\begin{align*}\int\tan^62x~dx&=\int\tan^22x\tan^42x~dx\\\\
&=\int(\sec^22x-1)\tan^42x~dx\\\\
&=\int\sec^22x\tan^42x~dx-\int\tan^42x~dx\\\\
&=\int\sec^22x\tan^42x~dx-\int\tan^22x\tan^22x~dx\\\\
&=\int\sec^22x\tan^42x~dx-\int(\sec^22x-1)\tan^22x~dx\\\\
&=\int\color{blue}{\sec^22x}\color{red}{\tan}^4\color{red}{2x}~\color{blue}{dx}-\int\sec^22x\tan^22x~dx+\int\tan^22x~dx\\\\
&=\int\sec^22x\tan^42x~dx-\int\sec^22x\tan^22x~dx+\int\sec^22x~dx\\
&\quad \quad-\int dx

\end{align*}\]
Substitute \(\color{red}u=\color{red}{\tan2x}\), which would give \(\color{blue}{du}=\color{blue}{2\sec^22x~dx}\), then you have
\[\frac{1}{2}\int u^4~du-\frac{1}{2}\int u^2~du+\int\sec^22x~dx-\int dx\]