Trig Substitution: int [x^2/(x^2 - 8)^(5/2)] dx

[Sga001]

Hello, I was given the following equation and I was asked to integrate it without using hyperbolic trig functions, I tried by many different ways but couldn't.

I would highly appreciate if someone could point me in the right track?

\(\displaystyle \L\int {{\textstyle{{x^2 } \over {(x^2 - 8)^{{\textstyle{ \over 5 2}}} }}}}dx\)

Here is what i'm doing: replacing x for sec(x) and then going from there.

 

[soroban]

Re: Trig Substitution

Hello, Sga001!

I was given the following to integrate
without using hyperbolic trig functions.

\(\displaystyle \L \int \frac{x^2}{(x^2\,-\,8)^{\frac{5}{2}}}\,dx\)

Let \(\displaystyle x \:=\:\sqrt{8}\cdot\sec\theta\;\;\Rightarrow\;\;dx\:=\:\sqrt{8}\cdot\sec\theta\tan\theta\,d\theta\)

. . Then: \(\displaystyle \:x^2\,-\,8\:=\:8\cdot\sec^2\theta\,-\,8\:=\:8(\sec^2\theta\,-\,1)\:=\:8\cdot\tan\theta\)


Substitute: \(\displaystyle \L\:\int\frac{(\sqrt{8}\cdot\sec\theta)^2}{\left(8\cdot\tan^2\theta\right)^{\frac{5}{2}}}\) \(\displaystyle \left(\sqrt{8}\cdot\sec\theta\tan\theta\,d\theta \right)\)

. . \(\displaystyle \L=\;\int\frac{8\cdot\sec^2\theta}{8^{\frac{5}{2}}\cdot\tan^5\theta}\) \(\displaystyle \left(\sqrt{8}\cdot\sec\theta\tan\theta\,d\theta\right)\L \;=\;\frac{1}{8}\int\frac{\sec^3\theta}{\tan^4\theta}\,d\theta\)

. . \(\displaystyle \L=\;\frac{1}{8}\int \frac{\:\frac{1}{\cos^3\theta}\:}{\frac{\sin^4\theta}{\cos^4\theta}}\,d\theta \;=\;\frac{1}{8}\int\)\(\displaystyle \left(\sin\theta)^{-4}\,(\cos\theta\,d\theta\right)\)


Let \(\displaystyle \L u \:=\:\sin\theta\)


Can you finish it?