- #1
karush
Gold Member
MHB
- 3,269
- 5
$\text{206.8.7.32}$
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?
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?