Simple Trigonometric Substitution

[tangibleLime]

Homework Statement

$$\int \frac{4}{x^{2}\sqrt{81-x^{2}}} dx$$

Homework Equations

The Attempt at a Solution

Since the radical is of the form $$a^2-x^2$$, I'm using the substitution $$x=asin\theta$$.

$$x = 9sin\theta$$

$$dx = 9cos\theta d\theta$$​

Using this x value, I solved the radical and use the trig identity to replace 1-sin^2 with cos^2.

$$\sqrt{81-x^{2}}$$

$$\sqrt{81 - (9sin\theta)^{2}}$$

$$\sqrt{81(1-sin^2\theta)}$$

$$\sqrt{81cos^2\theta)}$$

$$9cos\theta$$​

Then I threw everything back into my original integral.

$$\int \frac{36cos\theta}{81sin^2\theta9cos\theta} d\theta$$​

Canceling and simplifying...

$$\int \frac{4cos\theta}{81sin^2\theta} d\theta$$

This is where I get lost. I don't think I'm on the right track. I've watched several demonstrations of this kind of problem, and they all work out much better than this. Usually, I think, because there's a 1 on top instead of a 4. Any hints would be great.

 

[HallsofIvy]

?? 4 instead of 1? They are both constants- take them out of the integral!

What you have at the end is a standard, simple integral- it has an odd power of cosine.

Let $u= sin(\theta)$. Then $du= cos(\theta)d\theta$ and your integral becomes
$$\frac{4}{81}\int u^{-2}du$$

 

[tangibleLime]

Oh cripes, thanks. This is what starts happening when I don't sleep 0_0

 

[betel]

I don't see how you get from your second last to your last line. Why didn't you cancel the $$\cos\theta$$
Otherwise it would be very strange. By changing variables to sin and back you get rid of the root with nothing but scaling.

And if you want to integrate $$\int\frac{1}{\sin^2\theta}d\theta$$. There is an easy antiderivative for this.