Math (Cal)
posted by Angel on .
I having difficulty integrating this
Ingrate (x^3)/sqrt(x^2  9)
I know that x= 3sec(theta)
Sqrt(x^29)= 3tan(theta)
x^3 = 27Sec^3(theta)
Once I put that all together I'm not sure what to do next.

so far, so good. But you forgot about dx:
dx = 3secθtanθ dθ
So, now you have
∫27sec^3(θ) / 3tanθ * 3secθtanθ
= 27∫sec^4(θ) dθ
= 27∫(tan^2(θ)+1) sec^2(θ) dθ
Now if you let u = tanθ, you have
= 27∫(u^2+1) du
= 27(u^3/3 + u)
= 27(tan^3(θ)/3 + tanθ)
= 9tanθ (tan^2(θ) + 3)
Now, the original substitutions,
tanθ = 1/3 √(x^29), so we wind up with
3√(x^29) (1/9 (x^29) + 3)
= 1/3 √(x^29) (x^29+27)
= 1/3 √(x^29) (x^2+18)