# Trigonometric Substitution Problem w/ Sin Substitution

1. Nov 19, 2016

### Burjam

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

∫(√(64 - x^2)) / x dx

I must solve this using a sin substitution.

2. Relevant equations

x = 8sinΘ
dx = 8cosΘ dΘ
Θ = arcsin(x/8)
Pythagorean Identities

3. The attempt at a solution

(After substitution)
= ∫8cosΘ * (√(64 - 64sin^2Θ)) / 8sinΘ dΘ
= ∫(cosΘ * (√(64(1 - sin^2Θ))) / sinΘ dΘ
= ∫(cosΘ * (√(64cos^2Θ)) / sinΘ dΘ
= ∫8cos^2Θ / sinΘ dΘ

At this point, I didn't see any easy substitutions, like the rest of the problem set. So I thought there was a good chance I made an error. I checked to see if the answer to this integral was the same as the answer to the original integral, and saw that it wasn't. So somewhere up to this point, I made a mistake. After looking over my work, I don't see what I did wrong. Maybe it's just the integral calculator I'm using, I'm not sure.

Either way, I don't really know how to proceed with this problem. Sorry if the formatting is hard to read. I don't know the bbcode for posting these symbols properly and am in a rush at the moment.

2. Nov 19, 2016

### Simon Bridge

This reads: $$\int \frac{\sqrt{64-x^2}}{x}\; dx$$
... is that correct?

If so, then sub $x=8\sin\theta$ does get you $dx = 8\cos\theta\; d\theta$
To give: $$8\int \frac{\cos^2\theta}{\sin\theta}\; d\theta$$ ... which is as far as you've got.

You need to use trig identities to turn this into integrals you know how to do ... how about $\cos^2\theta = 1-\sin^2\theta$ ?
The integrand is also $\cos\theta / \tan\theta$ and other forms you may have some luck figuring out.
Basically you need to find a big table of trig identities and integrals.

3. Nov 19, 2016

### Burjam

That is correct. And thank you, I was able to work out the solution. I don't know why, but I kept insisting to myself that this simplified integral needed a u substitution when in reality all I had to do was simplify it further with some trig identities. Sometimes when I do problems in a rush I make mistakes like this.