# Integration via Trigonometric Substitution

## 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.

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

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

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.

So what is wrong with this:
$$\int{\frac{sin^{2}θ}{(cos^{2}θ)^\frac{5}{2}}}dθ= Oh, I misunderstand what you want to know. I thought you want to calculate the last thing. When you starts from the beginning, there should be ## dx##, the integral variable. Then, I think you may find what you missed during your calculation. Yeah I forgot to type that in, anyway it's trig identity I'm kinda having trouble with atm. ## dx ## cannot change directly ## d\theta ##. They have to connected by some function. ## dx ## cannot change directly ## d\theta ##. They have to connected by some function. Not too sure what you mean by that. For functions with the form [tex]\sqrt{a^2-x^2}$$ you can express them as $$x=asinθ$$

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.

Yeah it was a typo...

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

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

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 ##.

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