Help with trigonometric substitution

[ronybhai]

Homework Statement

integral (1)/(x^2sqrt(36-x^2)

Homework Equations

The Attempt at a Solution

I found X=6sinθ dx=6cos
√(36-x^2)=√(36-sin^2θ)=6cosθ
i think the problem is that i am not getting integral of ∫csc^2θ

 

[Dick]

Homework Statement

integral (1)/(x^2sqrt(36-x^2)

Homework Equations

The Attempt at a Solution

I found X=6sinθ dx=6cos
√(36-x^2)=√(36-sin^2θ)=6cosθ
i think the problem is that i am not getting integral of ∫csc^2θ

You should. Show the rest of your work so someone can tell you where you went wrong.

 

[ronybhai]

You should. Show the rest of your work so someone can tell you where you went wrong.

i know the answer which is -√(36-x^2)/36x+C

 

[Dick]

i know the answer which is -√(36-x^2)/36x+C

That's not the rest of your work, that's what I asked for. You've got a cos(θ) and the numerator and cos(θ) in the denominator. They cancel. What's left?

 

[KDeep]

i know the answer which is -√(36-x^2)/36x+C

The integral is -cotθ + C

 

[Dick]

The integral is -cotθ + C

That's part of it. There's also a constant around. But like we were talking about in your last post there is a way to express -cot(arcsin(x/6)) as a function of x without any trig functions. That's what the books answer is.