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.
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!
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