
Yefim S. answered 02/28/21
Math Tutor with Experience
x = 2secθ; dx = 2secθtanθdθ; Then ∫x^3√x^2 − 4 dx
= ∫8sec3θ·2tanθ·2secθtanθdθ = 32 ∫sec4θtan2θdθ = 32 ∫(tan2 + 1)tan2θd(tanθ) = 32(tan5θ/5 + tan3θ/3) + C;
tanθ = √x2 - 4/2 So, our integral I = (√x2 - 4)5/5 + 4(√x2 - 4)3/3 + C