Question

Integral rational trigonometric ​

Answer 1

Substitute x = 3 - 2 cos(θ) and dx = 2 sin(θ) dθ (where "sin" = "sen"). So we have

∫ sin(θ) / (3 - 2 cos(θ)) dθ = 1/2 ∫ 1/x dx

= 1/2 ln|x| + C

= 1/2 ln(3 - 2 cos(θ)) + C

(We can remove the absolute value because -1 ≤ cos(θ) ≤ 1, so 1 ≤ 3 - 2 cos(θ) ≤ 5, and |x| = x when x ≥ 0.)