Integrating a trigonometric integral using trig identities (calculus 2)

Question

Evaluate the following integral

∫sin⁶(x)cos⁵(x)dx

Answer 1

Recall that sin²x + cos²x = 1. So, cos²x = 1 - sin²x.

∫sin⁶xcos⁵xdx = ∫sin⁶xcos⁴xcosxdx = ∫sin⁶x (1 - sin²x)²cosxdx

Let u = sinx. Then du = cosxdx.

So, we have ∫ u⁶(1 - u²)²du = ∫ u⁶ (1 - 2u² + u⁴)du = ∫ [u⁶ - 2u⁸ + u¹⁰]du

= (1/7)u⁷ - (2/9)u⁹ + (1/11)u¹¹ + C = (1/7)sin⁷x - (2/9)sin⁹x + (1/11)sin¹¹x + C

Answer 2

Video on integrating trigonometric integral by u-substitution and trigonometric identities!!