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.