Integrate tan^5(x/4)dx with trig identities and substitution

[quazzi]

Can't get this (tan(x/4))^5dx problem, looks like it would use the power reducing formula but get even more screwes up. help appreciated.

 

[pka]

\(\displaystyle \begin{array}{rcl}
\tan ^5 (u) & = & \left( {\sec ^2 (u) - 1} \right)^2 \tan (u) \\
& = & \left( {\sec ^4 (u) - 2\sec ^2 (u) + 1} \right)\tan (u) \\
\end{array}\)
The last expression is easy to intergrate.

 

[soroban]

Re: Integrate tan^5(x/4)dx with trig identities and substitu

Hello, quazzi!

\(\displaystyle \L\int\tan^5\left(\frac{x}{4}\right)\,dx\)

I would do it like this . . . Let \(\displaystyle \theta\,=\,\frac{x}{4}\)

\(\displaystyle \tan^5\theta \;=\;\tan^4\theta\cdot\tan\theta\)

. . . . . .\(\displaystyle = \;\left(\tan^2\theta\right)^2\tan\theta\)

. . . . . .\(\displaystyle = \;\left(\sec^2\theta\,-\,1\right)^2\tan\theta\)

. . . . . .\(\displaystyle = \;\left(\sec^4\theta\,-\,2\sec^2\theta\,+\,1\right)\tan\theta\)

. . . . . .\(\displaystyle = \;\sec^4\theta\cdot\tan\theta\,-\,2\sec^2\theta\cdot\tan\theta\,+\,\tan\theta\\\)

pka did all this . . . now I'll complete the solution.


We have: \(\displaystyle \:\L\int\sec^4\theta\cdot\tan\theta\,d\theta \:-\:2\int\sec^2\theta\cdot\tan\theta\,d\theta \,+\,\int\tan\theta\,d\theta\)


For the first integral, we have: \(\displaystyle \L\:\int\sec^3\theta(\sec\theta\cdot\tan\theta\,d\theta)\)
Let \(\displaystyle u \:=\:\sec\theta\)

For the second integral, we have: \(\displaystyle \L\:\int\tan\theta(\sec^2\theta\,d\theta)\)
Let \(\displaystyle u \:=\:\tan\theta\)

For the third integral, a formula: \(\displaystyle \L\:\int\tan\theta\,d\theta\:=\:-\ln|\cos\theta|\,+\,C\)