Question

1. integral (x^3 * sqrt(x^2-4)) dx

I'm confused on how to start this problem, if I am doing trig substitution.

Answer 1

No need for trig substitution. We'll just do normal substitution:
Let u = sqrt(x^2 - 4)
Thus, we can make the following relation:
u^2 = x^2 - 4
x^2 = u^2 + 4

2u du = 2x dx
dx = u/x du

Substituting,
∫ ( x^3 * sqrt(x^2-4) ) dx
∫ ( x^3 * u * u / x ) du
∫ ( x^2 * u^2 ) du
∫ ( (u^2 + 4) u^2 ) du
∫ ( u^4 + 4u^2 ) du
= (1/5)u^5 + (4/3)u^3 + C
Substituting back the expression for u:
= (1/5) [sqrt(x^2 - 4)]^5 + (4/3) [sqrt(x^2-4)]^3 + C
= (1/5)(x^2 - 4)^5/2 + (4/3)(x^2-4)^3/2 + C

You can further simplify this if we want, by first factoring the common factor, (x^2-4)^(3/2):
= (x^2-4)^(3/2) * [ (1/5)*(x^2-4) + 4/3 ] + C
Now, do the rest if you want to continue this simplification.

hope this helps~ `u`