# Trigonometric Substitution Problem w/ Sin Substitution

• Burjam
In summary, the problem requires solving the integral of (√(64 - x^2)) / x dx using a sin substitution. After substituting, the integrand simplifies to 8cos^2Θ / sinΘ. However, this can be further simplified using trig identities to 8∫(1-sin^2Θ) / sinΘ dΘ. By using trig identities and recognizing the integrand as a combination of known integrals, the solution can be found.
Burjam

## Homework Statement

∫(√(64 - x^2)) / x dx

I must solve this using a sin substitution.

## Homework Equations

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

## 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.

∫(√(64 - x^2)) / x dx

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.

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.

Simon Bridge

## 1. What is Trigonometric Substitution?

Trigonometric substitution is a technique used in calculus to simplify integrals involving square roots and trigonometric functions. It involves substituting a trigonometric expression in place of a variable in the original integral.

## 2. When do I need to use Trigonometric Substitution?

Trigonometric substitution is typically used when the integral involves a square root of a quadratic function or when the integral involves a combination of trigonometric functions.

## 3. How do I choose the correct trigonometric substitution?

The choice of trigonometric substitution depends on the form of the integral. For square root expressions, a substitution of the form x = a sin θ or x = a cos θ is used. For integrals involving trigonometric functions, the substitution of tan θ or sec θ can be used.

## 4. What is the Sin Substitution method in Trigonometric Substitution?

Sin substitution is a specific type of trigonometric substitution where the variable x is replaced with a sin θ expression. This method is used when the integral involves a quadratic function under a square root and can be simplified by using the identity sin^2 θ + cos^2 θ = 1.

## 5. Can Trigonometric Substitution be used for all integrals?

No, Trigonometric Substitution is a specific technique used for certain types of integrals involving trigonometric functions. It is not applicable to all integrals and other integration techniques may need to be used for different types of integrals.

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