How to Solve Integrals Using Trig Substitution?

[KDeep]

Homework Statement

integral of
dx/((9-(x^2))^(3/2)) A = 0, B = 3/2

Homework Equations

Trigonometry Substitutions



3. The Attempt at a Solution : I am stuck with this question. So far, I got
(1/9)integral of (1/cos^2(θ)) dθ

 

[haruspex]

Do you know a trig function with derivative sec-squared?

 

[KDeep]

That's my problem. My trigonometry is very limited. I do not know of a Anti-derivative of sec^2θ

 

[Dick]

That's my problem. My trigonometry is very limited. I do not know of a Anti-derivative of sec^2θ

What is the derivative of tan(θ)?

 

[KDeep]

the derivative of tan(∅) is sec^2(∅). I get 1/9(tan((sin(∅))) | , Could someone solve the entire problem for me ? and reply with the steps to get there.
This would be a lot of help!

 

[Dick]

the derivative of tan(∅) is sec^2(∅). I get 1/9(tan((sin(∅))) | , Could someone solve the entire problem for me ? and reply with the steps to get there.
This would be a lot of help!

That's not right. It's just a little confused. What you need to do is show how you got there and then people can tell you where you are going wrong.

 

[KDeep]

A = 0 , B = 3/2 ∫ dx/((9-x²)^3/2)

First I substituted , x = 3sin∅ , dx = 3cos∅d∅

∫ 3cos∅d∅/(3²-(3sin∅)²)^3/2

Then I factored.

∫ 3cos∅d∅/(3²(1-sin²∅))^3/2

Then I used the trig identity to convert 1-sin²∅ to cos²θ

∫ 3cos∅d∅/(3²cos²∅)^3/2

Then I simplified

(1/9) ∫ (1/cos²θ) d∅

Then I anti-derived that into..

(1/9) tanθ | [ I do not know what to do after this, Since it is respect to ∅]

To conclude, I cannot use A and B, Since those are respect to x. I am stuck on this part.

 

[ammus]

differential equation

Can you tell me how to crack the following problems?:

1) y=2x(dy/dx)-y(dy/dx)2

2) (d2y/dx2)+y=2 cos2(x)+sin(3x)

 

[Dick]

A = 0 , B = 3/2 ∫ dx/((9-x²)^3/2)

First I substituted , x = 3sin∅ , dx = 3cos∅d∅

∫ 3cos∅d∅/(3²-(3sin∅)²)^3/2

Then I factored.

∫ 3cos∅d∅/(3²(1-sin²∅))^3/2

Then I used the trig identity to convert 1-sin²∅ to cos²θ

∫ 3cos∅d∅/(3²cos²∅)^3/2

Then I simplified

(1/9) ∫ (1/cos²θ) d∅

Then I anti-derived that into..

(1/9) tanθ | [ I do not know what to do after this, Since it is respect to ∅]

To conclude, I cannot use A and B, Since those are respect to x. I am stuck on this part.

Pretty good. You've got both thetas and phis floating around there. Intended to mean the same thing, I hope. One way to do it is the find the theta limits, since you substituted x=3*sin(θ) you just need to solve 3/2=3*sin(θ) and 0=3*sin(θ). If you want to eliminate the trig functions altogether you can solve for theta using the arcsin function. θ=arcsin(x/3). There's also a way to simplify tan(arcsin(x/3)). Any of this sounding familiar?

 

[Dick]

Can you tell me how to crack the following problems?:

1) y=2x(dy/dx)-y(dy/dx)2

2) (d2y/dx2)+y=2 cos2(x)+sin(3x)

That has nothing to do with this thread. You really need to start a new one.

 

[KDeep]

Ok that's what I was going for.

So tan(sin^-1(x/3))

(sin^-1((3/2)/3)) is ∏/6 right?


1/9 (Tan (∏/6) - Tan(0)) ? Am I on the right track?

 

[Dick]

Ok that's what I was going for.

So tan(sin^-1(x/3))

(sin^-1((3/2)/3)) is ∏/6 right?1/9 (Tan (∏/6) - Tan(0)) ? Am I on the right track?

That's exactly right. It's also good practice to think about how you might express tan(θ) in terms of x. Since sin(θ)=x/3 and sine=opposite/hypotenuse draw a right triangle with opposite side length x and hypotenuse length 3. Then figure out the length of the adjacent side. Now tan(θ)=opposite/adjacent.

 

[KDeep]

Yeah thanks for the tip.

So the answer is 1/(9√3)? Am I correct?

 

[Dick]

Yeah thanks for the tip.

So the answer is 1/(9√3)? Am I correct?

Yes, it is.