How do I do this integral? int{0,1} [(-36)/(sqrt[36x^2+9])] dx

[math]\int_0^1 \dfrac{-36}{\sqrt{36x^2+9}} ~ dx = \int_0^1 \dfrac{-36}{3\sqrt{4x^2+1}} ~ dx[/math]
Does that remind you of the derivative of a particular trig function?
 
? There's a link in the forum's posting guidelines for how to text math expressions. We can type the integral like this:

int[0,1] -36/sqrt(36x^2+9) dx
[imath]\;[/imath]
 
No one here is going to solve this for you as that would not be helpful.
You were given a hint in post 2, did you get anywhere with that hint.
Can we see the work you tried so we know how to guide you?

I'll give you another hint. Factor out the constant and let u=2x
 
I already got to the final result. I just dont know why I need to do u=2x, for me it was u=4x. Can someone explain this step pls ?
 
2x is the principal square root of 4x^2, that's why it is a more appropriate substitution.
 
skarner32 said:
I already got to the final result. I just dont know why I need to do u=2x, for me it was u=4x. Can someone explain this step pls ?
Click to expand...
Take the derivate of your final answer and it will not give you back the integrand OR you made a mistake that cancelled out the error letting u= 4x

If u=2x, then u^2 = (2x)^2 = 4x^2. So 4x^2 + 1 becomes u^2 + 1
 
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