How to evaluate the Trigonometric Integrals : ∫ sin^2(1/3 θ) dθ , the upper limit is 2π and the lower limit is 0 ?

Question

Emilia Aguilar · Accepted Answer

Use the power reducing formula.

From #cos(2t) = 1-2sin^2t#, we get #sin^2t = 1/2(1-cos(2t))#. #int sin^2t dt = 1/2 int (1-cos(2t)) dt#, which is straightforward to integrat by substitution. In this question, #t = 1/3theta#, so we get #int sin^2(1/3theta) d theta = 1/2 int (1-cos(2/3theta)) d theta#. Integrate #1 d theta#, then #cos(2/3theta) d theta# by subsitution. Note We didn't need it here, but. also from #cos(2t) = 2cos^2t-1#, we get the power reduction for cosine. #cos^2t = 1/2(cos(2t)-1)#

Cameron Alexander · Answer

undefined To evaluate the integral ∫ sin²(1/3 θ) dθ from 0 to 2π, we can use the trigonometric identity sin²(θ) = 1/2 - 1/2 cos(2θ):

∫ sin²(1/3 θ) dθ = ∫ (1/2 - 1/2 cos(2(1/3 θ))) dθ
                  = ∫ (1/2 - 1/2 cos(2/3 θ)) dθ
                  = (1/2)θ - (1/4)sin(2/3 θ) + C

Evaluate this expression from 0 to 2π:

[(1/2)(2π) - (1/4)sin(2/3 (2π))] - [(1/2)(0) - (1/4)sin(2/3 (0))]
= π - 0
= π