Integration via Trigonometric Substitution

[Cpt Qwark]

Homework Statement

Evaluate $$\int{\frac{x^2}{(1-x^2)^\frac{5}{2}}}dx$$ via trigonometric substitution.
You can do this via normal u-substitution but I'm unsure of how to evaluate via trigonometric substitution.

Homework Equations

The Attempt at a Solution

Letting $$x=sinθ$$,
$$\int{\frac{sin^{2}θ}{(1-sin^{2}θ)^\frac{5}{2}}}dθ=\int{\frac{sin^{2}θ}{(cos^{2}θ)^\frac{5}{2}}}dθ$$
but I'm not sure how the working in the answers gets up to $$\int{\frac{x^2}{(1-x^2)^\frac{5}{2}}}dx=\int{\frac{sin^{2}θ}{cos^{4}θ}}dθ$$.

 

[Daeho Ro]

Try to change the last integral by using other trigonometric functions, like $\tan\theta$ and $\sec\theta.$

 

[Daeho Ro]

Oh, I misunderstand what you want to know. I thought you want to calculate the last thing.

When you starts from the beginning, there should be $dx$, the integral variable. Then, I think you may find what you missed during your calculation.

 

[Cpt Qwark]

So what is wrong with this:
[tex]\int{\frac{sin^{2}θ}{(cos^{2}θ)^\frac{5}{2}}}dθ=<br /> <blockquote data-attributes="member: 573183" data-quote="Daeho Ro" data-source="post: 5253790" cite="https://www.physicsforums.com/goto/post?id=5253790" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Daeho Ro said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Oh, I misunderstand what you want to know. I thought you want to calculate the last thing.<br /> <br /> When you starts from the beginning, there should be $dx$, the integral variable. Then, I think you may find what you missed during your calculation. </div> </div> </blockquote><br /> Yeah I forgot to type that in, anyway it's trig identity I'm kinda having trouble with atm.[/tex]

 

[Daeho Ro]

$dx$ cannot change directly $d\theta$. They have to connected by some function.

 

[Cpt Qwark]

$dx$ cannot change directly $d\theta$. They have to connected by some function.

Not too sure what you mean by that.
For functions with the form $$\sqrt{a^2-x^2}$$ you can express them as $$x=asinθ$$

 

[Daeho Ro]

See,

$$\int \dfrac{x^2}{(1-x^2)^{5/2}} dx = \int \dfrac{ \sin^2\theta}{(1-\sin^2\theta)^{5/2}} dx = \int \dfrac{\sin^2 \theta}{\cos^5\theta} dx \neq \int \dfrac{\sin^2 \theta}{\cos^5\theta} d\theta.$$
You missed something in the lase step.

 

[Cpt Qwark]

Yeah it was a typo...

 

[Daeho Ro]

Then, what is $dx$ as a function of $\theta$?

 

[Cpt Qwark]

The denominator is somehow supposed to be $$cos^{4}θ$$, not $$cos^{5}θ$$. That's all I need help with, nothing else.

 

[Daeho Ro]

The denominator is somehow supposed to be $$cos^{4}θ$$, not $$cos^{5}θ$$. That's all I need help with, nothing else.

Yes, I know and you almost reach the final goal.

$dx$ have to change as $d\theta$ because the last integration is in a form $\int (\cdots) d\theta$. But as you know, $dx \neq d\theta$.

What is $dx / d\theta$? It's really strong hint about this problem.

 

[Daeho Ro]

I hope you already got the answer for the problem.

The key idea is chain rule. The differentiation of $x$ with respect to $\theta$ is $dx/d\theta = \cos\theta$. Then, the integration will change as
$$ \int \dfrac{\sin^2\theta}{\cos^5\theta} dx = \int \dfrac{\sin^2\theta}{\cos^5\theta} \dfrac{dx}{d\theta} d\theta = \int \dfrac{\sin^2\theta}{\cos^5\theta} \cos \theta d\theta = \int \dfrac{\sin^2\theta}{\cos^4\theta} d\theta.$$