Section8_2_review

Section8_2_review - Section 8.2: Trigonometric Integrals...

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Section 8.2: Trigonometric Integrals Please refer to handout for review material. Example. Evaluate cos 5 xdx . Solution. cos 5 xdx = cos 4 x cos xdx = 1 - sin 2 x ( 29 2 cos xdx = 1 - 2sin 2 x + sin 4 x ( 29 cos xdx = cos xdx - 2sin 2 x cos xdx + sin 4 x cos xdx The first integral, cos xdx = sin x For the second and third integral, u = sin x du = cos xdx So cos 5 xdx = sin x - 2 u 2 du + u 4 du = sin x - 2 u 3 3 + u 5 5 + c = sin x - 2 3 sin 3 x + 1 5 sin 5 x + c Example. Evaluate sin 2 x cos 4 xdx . Solution. Using the identities: sin 2 x = 1 2 1 - cos2 x ( 29 cos 2 x = 1 2 1 + cos2 x ( 29 so cos 4 x = cos 2 x ( 29 2 = 1 2 1 + cos2 x ( 29 2 = 1 4 1 + 2cos2 x + cos 2 2 x ( 29 Therefore: sin 2 x cos 4 xdx = 1 2 1 - cos2 x ( 29 1 4 1 + 2cos2 x + cos 2 2 x ( 29 dx = 1 8 1 + 2cos2 x + cos 2 2 x - cos2 x - 2cos 2 2 x - cos 3 2 x ( 29 dx = 1 8 1 + cos2 x - cos 2 2 x - cos 3 2 x ( 29 dx = 1 8 dx + cos2 xdx - cos

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This note was uploaded on 01/12/2010 for the course MATH 44245 taught by Professor Famiglietti during the Fall '07 term at UC Irvine.

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Section8_2_review - Section 8.2: Trigonometric Integrals...

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