All right....continuation of an earlier problem.?

Question

Hey everybody...I'm still stuck on trig sub integrals.

I have integral sqrt (x^2-9)/ x^3 dx

and I've gotten down to

1/3 integral 1 - cos 2 theta .... but I'm stuck after that...
I know that theta is inverse sec (x/3). but I'm clueless of how to end this problem. If possible, please show me the steps of how you proceed through the problem.

Thank you.

Answer 1

∫√(x²-9)/x³ dx

Let x=3 sec θ, dx=3 sec θ tan θ dθ

3 ∫√(9 sec² θ-9)/(27 sec³ θ) sec θ tan θ dθ
1/9 ∫√(9 tan² θ)/sec² θ tan θ dθ
1/3 ∫tan² θ/sec² θ dθ
1/3 ∫tan² θ cos² θ dθ
1/3 ∫sin² θ dθ
1/3 ∫(1-cos (2θ))/2 dθ
1/6 ∫1 dθ - 1/6 ∫cos (2θ) dθ
θ/6 - 1/12 sin (2θ) + C
θ/6 - 1/6 sin θ cos θ + C
θ/6 - 1/6 √(1-cos² θ) cos θ + C
θ/6 - 1/6 √(1-1/sec² θ) 1/sec θ + C
θ/6 - 1/2 √(1-9/(9 sec² θ)) 1/(3 sec θ) + C
1/6 arcsec (x/3) - 1/(2x) √(1-9/x²) + C