K katyt New member Joined Sep 7, 2005 Messages 13 Oct 11, 2005 #1 I'm having some trouble integrating this: (Integral of) (sin(3pix))^2 from 0 to 1/3 I keep getting 0, but when I put it into my calculator, I get 1/6. Could someone give me some help?
I'm having some trouble integrating this: (Integral of) (sin(3pix))^2 from 0 to 1/3 I keep getting 0, but when I put it into my calculator, I get 1/6. Could someone give me some help?
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Oct 11, 2005 #2 Hello, Katy! Just a guess . . . you're probably integrating incorrectly. . . It's tricky . . . [INT] sin<sup>2</sup>(3 pi x) dx . . . from 0 to 1/3 . Click to expand... To integrate sine-squared, we use: .sin<sup>2</sup>θ = (1/2) [1 - cos(2θ)] . . So: .sin<sup>2</sup>(3πx) .= .(1/2) [1 - cos(6πx)] We are integrating: .(1/2) [INT] [1 - cos(6πx)] dx . . and we get: .(1/2) [x - (1/6π)sin(6πx) ] Now plug in the limits . . .
Hello, Katy! Just a guess . . . you're probably integrating incorrectly. . . It's tricky . . . [INT] sin<sup>2</sup>(3 pi x) dx . . . from 0 to 1/3 . Click to expand... To integrate sine-squared, we use: .sin<sup>2</sup>θ = (1/2) [1 - cos(2θ)] . . So: .sin<sup>2</sup>(3πx) .= .(1/2) [1 - cos(6πx)] We are integrating: .(1/2) [INT] [1 - cos(6πx)] dx . . and we get: .(1/2) [x - (1/6π)sin(6πx) ] Now plug in the limits . . .
