MHB Evaluating $\int \tan^9(x) \sec^4(x) dx$

[karush]

$\text{206.8.7.32}$
Given and evaluation
$$\displaystyle
I_{32}=\int \tan^9\left({x}\right)\sec^4(x) \, dx
=\dfrac{\tan^{12}\left(x\right)}{12}
+\dfrac{\tan^{10}\left(x\right)}{10} + C$$
use identity $\tan^2\left({x}\right)+1=\sec^2\left(x\right)$
$$u=\tan\left(x\right)
\therefore
du =\sec^{2}\left(x\right) \, dx$$
substitute and integrate
$$\displaystyle
I_{32}=\int u^9\left(u^2+1\right) \, du

\implies \int u^{11}+u^9 \, du
\implies \frac{u^{12}}{u^{12}}+\frac{u^{10}}{10} +C$$
backsubstute $u=\tan\left(x\right) $
$$I_{32}=\dfrac{\tan^{12}\left(x\right)}{12}
+\dfrac{\tan^{10}\left(x\right)}{10} + C $$Ok think I got this one ? Suggestions?

 

[HallsofIvy]

With me, it is almost a reflex to change everything to sine and cosine. tan= sine/cosine and secant= 1/cosine so tan^9(x)sec^4= \frac{sin^9(x)}{cos^9(x)}\frac{1}{cos^4(x)}= \frac{sin^9(x)}{cos^{13}(x)}. That has both sin and cosine to an odd power. Since sine is in the numerator, we can factor one sine out, to use with the differential, then change the even power of sine to cosine: \int \frac{sin^8(x)}{cos^{13}(x)}(sin(x)dx)= \int \frac{(1- cos^2(x))^4}{cos^{13}(x)}(sin(x)dx) Now, let u= cos(x) so that du= -sin(x)dx and that becomes -\int \frac{(1- u^2)^4}{u^{13}}du= -\int \frac{u^8- 4u^6+ 6u^4- 4u^2+ 1}{u^{13}}du= \int -u^{-5}+ 4u^{-7}- 6u^{-9}+4u^{-11}- u^{-13} du

 

[karush]

I think that's a good idea.
Was looking at this one terms of powers

 

[Greg]

$\text{206.8.7.32}$
Given and evaluation
$$\displaystyle
I_{32}=\int \tan^9\left({x}\right)\sec^4(x) \, dx
=\dfrac{\tan^{12}\left(x\right)}{12}
+\dfrac{\tan^{10}\left(x\right)}{10} + C$$
use identity $\tan^2\left({x}\right)+1=\sec^2\left(x\right)$
$$u=\tan\left(x\right)
\therefore
du =\sec^{2}\left(x\right) \, dx$$
substitute and integrate
$$\displaystyle
I_{32}=\int u^9\left(u^2+1\right) \, du

\implies \int u^{11}+u^9 \, du
\implies \frac{u^{12}}{u^{12}}+\frac{u^{10}}{10} +C$$
backsubstute $u=\tan\left(x\right) $
$$I_{32}=\dfrac{\tan^{12}\left(x\right)}{12}
+\dfrac{\tan^{10}\left(x\right)}{10} + C $$Ok think I got this one ? Suggestions?

Good work! Works fine! (Minus the typo - can you spot it)?

 

[karush]

typo no see