Evaluate the Integral. Show All Steps

Question

Evaluate the Integral. Show All Steps. If possible attach a video going through the problem.

Answer 1

Let us start by recalling some of the half angle formulas:

sin²θ = (1 - cos(2θ)) / 2

cos²θ = (1 + cos(2θ)) / 2

Let us also recall a quadratic identity that we will use:

(a - b)² = a² - 2ab + b²

Let us work with sin⁴θ for simplicity. We will replace the angle θ by (1/2)x at the end.

sin⁴θ = (sin²θ)²

= [(1 - cos(2θ)) / 2]² (here we used one of the half angle identities from above on sin²θ)

= (1 - cos(2θ))² / 2²

= (1 - 2cos(2θ) + cos²(2θ)) / 4 (here we used the quadratic identity from above)

= 1/4 - (2/4)cos(2θ) + (1/4) cos²(2θ)

= 1/4 - (1/2)cos(2θ) + (1/4) cos²(2θ)

= 1/4 - (1/2)cos(2θ) + (1/4) (1 + cos(2*2θ)) / 2 (we used a half angle identity again, this time on cos²(2θ))

= 1/4 - (1/2)cos(2θ) + (1/8) (1 + cos(4θ))

= 1/4 - (1/2)cos(2θ) + 1/8 + (1/8) cos(4θ)

= 1/4 + 1/8 - (1/2)cos(2θ) + (1/8) cos(4θ)

= 2/8 + 1/8 - (1/2)cos(2θ) + (1/8) cos(4θ)

= 3/8 - (1/2)cos(2θ) + (1/8) cos(4θ)

sin⁴θ = 3/8 - (1/2)cos(2θ) + (1/8) cos(4θ)

sin⁴(x/2) = 3/8 - (1/2)cos(2 * x/2) + (1/8) cos(4 * x/2) = 3/8 - (1/2)cos(x) + (1/8) cos(2x)

∫ sin⁴(x/2) dx

= ∫ [3/8 - (1/2)cos(x) + (1/8) cos(2x)] dx

= ∫ (3/8) dx - (1/2)∫cos(x) dx + (1/8) ∫ cos(2x) dx

= (3/8)x - (1/2)sin(x) + (1/16)sin(2x) + C

Detailed explanation of (1/8) ∫ cos(2x) dx

Let u = 2x.

Then du = 2dx so dx = du / 2

(1/8) ∫ cos(2x) dx = (1/8) ∫ cos(u) (du/2) = (1/8) * (1/2) ∫ cos(u) (du) = (1/16) sin(u) + C' = (1/16) sin(2x) + C'