Trigonometric Substitution

Question

Evaluate indefinte intergral of

(x^10)/((9-x^2)^(13/2))

thank you!!

Answer 1

The substitution x = 3 sin θ will get you (after some algebra) an integrand of (1/9) tan¹⁰θ sec²θ which integrates by the power rule,

Answer 2

∫x¹⁰/(9-x²)^13/2dx

Let x=3sin(θ) dx = 3cos(θ)dθ sin(θ) = x/3

∫3¹¹sin¹¹(θ)/(9^13/2(1-sin²(θ))^13/2 dθ = (3¹¹/3¹³)∫sin¹⁰(θ)cos(θ)/cos¹³(θ) dθ

= (1/3²)∫tan¹⁰(θ)sec²(θ) dθ

Let u = tan(θ) du = sec²(θ)dθ

(1/9)∫u¹⁰du = u¹¹/99 = tan¹¹(θ)/99 + C

If sin(θ)=x/3 then cos(θ) = √(9-x²)/3 and tan(θ) = x/√(9-x²)

∫x¹⁰/(9-x²)^13/2dx = x¹¹/(99(9-x²)^11/2) + C

Answer 3

Make substitution x=3sint, Then dx= costdt and integral turns out to be

3^11 ∫tany d(tant)

The rest should be easy, I guess.