Question

Hello! I have this problem:

x(dx)/sqrt(9-x^2)

I was wondering why I can't use trig substitution and substitute sqrt(9-x^2) for sqrt(1-sec^2) and having:
integral x = 3sin(theta)
dx = 3cos(theta)d(theata)

integral 3sin(theta)(3cos(theta))/3cos(theta)
having the 3cos(theta) cancel and leaving integral 3sin(theta) d(theta)
equalling 3cos(theta) and coming to an answer of sqrt(9-x^2)/3 + C?

Thank You!

Answer 1

you can do the trig stuff, but why bother?

let u = 9-x^2
du = -2x dx

and the integrand becomes

-1/2 u^(-1/2) du
and integrates to

- u^(1/2) = -√(9-x^2)

As for the trig, you cannot take √(1-sec^2) since sec^2 is always greater than 1, and you don't have a real value.

I'd so

x = 3sinθ
dx = 3cosθ dθ
√(9-x^2) = √(9-9sin^2θ) = 3cosθ

x/√(9-x^2) dx = 3sinθ/3cosθ * 3cosθ dθ
= 3sinθ dθ

the integral is then just

-3cosθ = -√(9-x^2) + C

You lost a factor of 3 in there somewhere.