Simple Trigonometric Substitution (1 Viewer)

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1. The problem statement, all variables and given/known data
[tex]\int \frac{4}{x^{2}\sqrt{81-x^{2}}} dx[/tex]

2. Relevant equations

3. The attempt at a solution

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

[tex]x = 9sin\theta[/tex]

[tex]dx = 9cos\theta d\theta[/tex]​

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


[tex]\sqrt{81 - (9sin\theta)^{2}}[/tex]




Then I threw everything back into my original integral.

[tex]\int \frac{36cos\theta}{81sin^2\theta9cos\theta} d\theta[/tex]​

Canceling and simplifying...

[tex]\int \frac{4cos\theta}{81sin^2\theta} d\theta[/tex]

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.


Science Advisor
?? 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 [itex]u= sin(\theta)[/itex]. Then [itex]du= cos(\theta)d\theta[/itex] and your integral becomes
[tex]\frac{4}{81}\int u^{-2}du[/tex]
Oh cripes, thanks. This is what starts happening when I don't sleep 0_0
I don't see how you get from your second last to your last line. Why didn't you cancel the [tex]\cos\theta[/tex]
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 [tex]\int\frac{1}{\sin^2\theta}d\theta[/tex]. There is an easy antiderivative for this.

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