Need help intergrating
1dx/[(4x − x^2)^3/2]
and
1dx/[x^2√(x^2-9)]
Thanks
This kind of stuff usually works best with a trig substitution.
Note that 4x-x^2 = 4-(x-2)^2
So, let u = x-2 and you have
du/(4-u^2)^(3/2)
Now, let
u = 2sinθ
du = 2cosθ dθ
4-u^2 = 4-4sin^2θ = 4cos^2θ
Now the integrand is
2cosθ dθ/8cos^3θ
= 4sec^2 θ dθ
which integrates to
4tanθ = 4sinθ/cosθ = 4u/√(4-u^2)
= (x-2)/√(4x-x^2)
The other one is a bit more complicated. Let
x = 3coshθ
x^2 = 9cosh^2 θ
x^2-9 = 9sinh^2 θ
dx = 3sinhθ dθ
Now the integrand becomes
3sinhθ dθ/(9cosh^2θ*3sinhθ)
= dθ/9cosh^2θ
= 1/9 sech^2θ dθ
which integrates to
1/9 tanhθ
= sinhθ/9coshθ
= √(x^2-9)/9x
You can also work with regular trig functions:
x = 3secθ
x^2 = 9sec^2 θ
x^2-9 = 9tan^2θ
dx = 3secθtanθ dθ
and you can take it from there.