# Integration via Trigonometric Substitution

Tags:
1. Oct 10, 2015

### Cpt Qwark

1. The problem statement, all variables and given/known data
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.
2. Relevant equations

3. 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 $$\int{\frac{x^2}{(1-x^2)^\frac{5}{2}}}dx=\int{\frac{sin^{2}θ}{cos^{4}θ}}dθ$$.

Last edited: Oct 10, 2015
2. Oct 10, 2015

### Daeho Ro

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

3. Oct 10, 2015

### 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.

4. Oct 10, 2015

### Cpt Qwark

So what is wrong with this:
$$\int{\frac{sin^{2}θ}{(cos^{2}θ)^\frac{5}{2}}}dθ= Yeah I forgot to type that in, anyway it's trig identity I'm kinda having trouble with atm. 5. Oct 10, 2015 ### Daeho Ro $dx$ cannot change directly $d\theta$. They have to connected by some function. 6. Oct 10, 2015 ### Cpt Qwark Not too sure what you mean by that. For functions with the form [tex]\sqrt{a^2-x^2}$$ you can express them as $$x=asinθ$$

7. Oct 10, 2015

### 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.

8. Oct 10, 2015

### Cpt Qwark

Yeah it was a typo...

9. Oct 10, 2015

### Daeho Ro

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

10. Oct 10, 2015

### 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.

11. Oct 10, 2015

### Daeho Ro

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.

12. Oct 10, 2015

### 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
$$\int \dfrac{\sin^2\theta}{\cos^5\theta} dx = \int \dfrac{\sin^2\theta}{\cos^5\theta} \dfrac{dx}{d\theta} d\theta = \int \dfrac{\sin^2\theta}{\cos^5\theta} \cos \theta d\theta = \int \dfrac{\sin^2\theta}{\cos^4\theta} d\theta.$$