q6-sol-1042-sp08

q6-sol-1042-sp08 - Calculus II-1042 Solution of Quiz # 6...

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Calculus II—1042 Solution of Quiz # 6 Fri., 03/07/08 1. Z π 0 8 cos 4 (2 x ) dx . Solution: Since cosine has EVEN exponent so break the even power using cos 2 x = 1 2 + cos(2 x ) 2 . We continue to use it till no even power left. First we do some algebra to break the even powers of cos 4 (2 x ) then we compute its anti-derivative 8 cos 4 (2 x ) = 8 ( cos 2 (2 x ) ) 2 = 8 ± 1 2 + cos(4 x ) 2 2 = 8 4 ( 1 + 2 cos(4 x ) + cos 2 (4 x ) ) = 2 ± 1 + 2 cos(4 x ) + 1 2 + cos(8 x ) 2 = 2 ± 3 2 + 2 cos(4 x ) + cos(8 x ) 2 = 3 + 4 cos(4 x ) + cos(8 x ) Z π/ 8 0 8 cos 4 (2 x ) dx = Z π/ 8 0 (3 + 4 cos(4 x ) + cos(8 x )) dx = 3 x + 4 4 sin(4 x ) + 1 8 sin(8 x ) | π/ 8 0 = 3 π 8 + sin( π 2 ) + 1 8 sin( π ) - ± 0 + sin(0) + 1 8 sin(0) = 3 π 8 + 1 2. Z sin 4 y cos 3 y dy . Solution: Since cos y has ODD power so split one cos y and convert the remaining cos 2 y into sin y using cos 2 y = 1 - sin 2 y , then make a substitution u = sin y Z sin 4 y cos 3 y dy = Z sin 4 y · cos 2 y · cos y dy = Z sin 4
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