using trig substitution evaluate the integral [x^3*square root(100+x^2)]dx

Question

[x^3•√(100+x^2)]dx

Answer 1

Do the substitution: x = 10tanθ dx = 10sec²θ dθ

integral of x³sqrt(100+x²)dx becomes integral of 10³tan³θsqrt(100+10²tan²θ)(10sec²θ)dθ

Which simplifies to 10⁵tan³θ sec³θ dθ which doesn't have a nice uⁿdu form with sec², but if we set aside sectan, we get

10⁵tan²sec²(sectan)dθ. Using tan² = sec²-1, we get an integrand which is secⁿ and its differential sectandθ

(I got tired of inserting θ.

Final integral is 10⁵(sec⁴-sec²)sectan dθ which is integrable as (u⁴ + u²)du

Your answer is in terms of powers of secθ which can be interpreted in x x/10 = tanθ, secθ = sqrt(x²+10²)/10

Take care.