This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Math 128: Trigonometric Integrals Koray Karabina ([email protected]) (Sep. 19, 2011) 1 Evaluating R sin m x cos n x dx 1.1 The power of cos x is odd: n = 2 k + 1 and k ≥ 1 Z sin m x cos 2 k +1 xdx = Z sin m x (cos 2 x ) k cos xdx = Z sin m x (1- sin 2 x ) k cos xdx (cos 2 x = 1- sin 2 x ) = Z u m (1- u 2 ) k du ( u = sin x, du = cos xdx ) = easy to compute; do not forget to substitute sin x = u at the end 1.2 The power of sin x is odd: m = 2 k + 1 and k ≥ 1 Z sin 2 k +1 x cos n xdx = Z (sin 2 x ) k sin x cos n xdx = Z cos n x (1- cos 2 x ) k sin xdx (sin 2 x = 1- cos 2 x ) =- Z u n (1- u 2 ) k du ( u = cos x, du =- sin xdx ) = easy to compute; do not forget to substitute cos x = u at the end 1.3 The power of sin x and cos x are even In this case, the above strategies will not work. You should use the half-angle identities in order to cut the power of trigonometric functions by half: sin 2 x = 1- cos2 x 2 cos 2 x = 1 + cos2 x 2 Typically, you will have to use the above identities more than once, and you may also need the...
View Full Document
- Fall '08
- Trigonometry, dx