# 7.2 - 7.2: Trigonometric Integrals (Dated: October 15,...

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7.2: Trigonometric Integrals (Dated: October 15, 2011) STRATEGY FOR EVALUATING R sin m x cos n x dx , WHERE m 0, n 0 ARE INTEGERS 1. If the power of cosine is odd ( n = 2 k + 1), save one co- sine factor and use cos 2 x = 1 - sin 2 x to express the remaining factors in terms of sine: Z sin m x cos 2 k + 1 x dx = Z sin m x (cos 2 x ) k cos x dx = Z sin m x (1 - sin 2 x ) k cos x dx . Then use the substitution u = sin x . 2. If the power of sine is odd ( m = 2 k + 1), save one sine factor and use sin 2 x = 1 - cos 2 x to express the remaining factors in terms of cosine: Z sin 2 k + 1 x cos n x dx = Z (sin 2 x ) k cos n x sin x dx = Z (1 - cos 2 x ) k cos n x sin x dx . Then use the substitution u = cos x . 3. If the powers of both sine and cosine are even, use the half-angle identities sin 2 x = 1 - cos2 x 2 and cos 2 x = 1 + cos2 x 2 . The identity sin x cos x = sin2 x 2 is sometimes helpful. Example 1 (Exercise 7.2.2 in the text) Z sin 6 x cos 3 x dx = Z sin 6 x cos 2 cos x dx = Z sin 6 x (1 - sin 2 x )cos x dx . Letting u = sin x yields du = cos x dx and thus Z sin 6 x cos 3 x dx = Z u 6 (1 - u 2 ) du = u 7 7 - u 9 9 + C = sin 7 x 7 - sin 9 x 9 + C . Example 2 (Exercise 7.2.3 in the text) Z sin 5 x cos 3 x dx = Z sin 4 x cos 3 x sin x dx = Z (sin 2 x ) 2 cos 3 x sin x dx = Z (1 - cos 2 x ) 2 cos 3 x sin x dx . Letting u = cos x yields - du = sin x dx and thus Z sin 5 x cos 3 x dx =- Z (1 - u 2 ) 2 u 3 du =- Z (1 - 2 u 2 + u 4 ) u 3 du =- u 4 4 + u 6 3 - u 8 8 + C =- cos 4 x 4 + cos 6 x 3 - cos 8 x 8 + C . Hence Z 3 π /4 π /2 sin 5 x cos 3 x dx = - cos 4 x 4 + cos 6 x 3 - cos 8 x 8 3 π /4 π /2 =- 11 384 . Example 3 (Exercise 7.2.3 in the text (revisited)) Z sin 5 x cos 3 x dx = Z sin 5 x cos 2 x cos x dx = Z sin 5 x (1 - sin 2 x )cos x dx . Letting u = sin x yields du = cos x dx and thus Z sin 5 x cos 3 x dx = Z u 5 (1 - u 2 ) du = u 6 6 - u 8 8 + C = sin 6 x 6 - sin 8 x 8 + C . Hence Z 3 π /4 π /2 sin 5 x cos 3 x dx = sin 6 x 6 - sin 8 x 8 3 π /4 π /2 =- 11 384 . Example 4 (Exercise 7.2.6 in the text) Letting u = p x yields 2 du = dx p x and thus Z sin 3 p x p x dx = 2 Z sin 3 u du . Now Z sin 3 u du = Z sin 2 u sin u dx = Z (1 - cos 2 u )sin u du .

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2 Letting v = cos u yields - dv = sin u du and Z sin 3 u du = Z ( v 2 - 1) dv = v 3 3 - v + C 1 = cos 3 u 3 - cos u + C 1 . Hence
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## This note was uploaded on 11/11/2011 for the course MATH 152.01 taught by Professor Geline during the Fall '09 term at Ohio State.

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7.2 - 7.2: Trigonometric Integrals (Dated: October 15,...

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