evaluate the integral

Question

evaluate the integral: (16x^2)/(sqrt(49-x^2))

the trig substitution for this is: 7sin(theta)

a. Which integral do you obtain after substituting for and simplifying?

b. What is the value of the above integral in terms of  ?

c. What is the value of the original integral in terms of ?

Answer 1

a. x = 7sinθ; dx = 7cosθdθ; ∫16x²dx/√(49 - x²) = ∫16·49sin²θ·7cosθdθ/(7cosθ) = ∫784sin²θdθ =

392∫(1 - coss2θ)dθ.

b. 392∫(1 - cos2θ)dθ = 392θ - 196sin2θ + C = 392(θ - sinθcosθ) + C

c. θ = sin^-1(x/7); sinθ = x/7; cosθ = √1 - x²/49 = √(49 - x²)/7;

∫16x²dx/√(49 - x²) = 392(sin^-1(x/7) - x/7·√(49 - x²)/7) + C = 392sin^-1(x/7) - 8x√(49 - x²) + C