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asm101W10-06sol - MATH 101 Assignment Date February 8 2009...

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MATH 101 - Assignment Date: February 8, 2009 In-class Assignment 06 Solutions The integral smorgasborg Closed book (except the integration guide line, last page) For each question compute one of the two given integrals - 2 hours. 1. (a) Compute the following indefinite integral ˆ e 2arccos x d x . We start with a substitution t = arccos x cos t = x , we get d x = - sin t d t . The integral becomes ˆ e 2arccos x d x = - ˆ e 2 t sin t d t which can be integrated by part. With - sin t d t = d v and e 2 t = u we have - ˆ e 2 t sin t d t = e 2 t cos t - 2 ˆ e 2 t cos t d t . We integrate by part a second time with cos t d t = d v and e 2 t = u we have ˆ e 2 t cos t d t = e 2 t sin t - 2 ˆ e 2 t sin t d t . Back substituting in the first integral yields - ˆ e 2 t sin t d t = e 2 t cos t - 2 e 2 t sin t + 4 ˆ e 2 t sin t d t and solving for - ´ e 2 t sin t d t gives - ˆ e 2 t sin t d t = 1 5 e 2 t cos t - 2 5 e 2 t sin t . 1
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MATH 101 Assignment KEYANO COLLEGE Back substituting t by arccos x and using the identity sin ( arccos x ) = 1 - x 2 we have the result ˆ e 2arccos x d x = 1 5 e 2arccos x x - 2 1 - x 2 + C . Note that this integral is not solved by Maple but is solved by Mathematica. (b) Compute the following indefinite integral ˆ e 2arcsinh x d x . Here we substitute ln x + 1 + x 2 for arcsin x . It yields ˆ e 2arcsinh x d x = ˆ e 2ln x + 1 + x 2 d x = ˆ e ln x + 1 + x 2 2 d x . With the cancellation equation, we get ˆ e 2arcsinh x d x = ˆ x + 1 + x 2 2 d x = ˆ x 2 + 2 x 1 + x 2 + 1 + x 2 d x . Collecting like terms and splitting gives ˆ e 2arcsinh x d x = ˆ d x + 2 ˆ x 2 d x + ˆ 1 + x 2 2 x d x The last integral is a ´ u d u type of integral when u = 1 + x 2 so that ˆ e 2arcsinh x d x = x + 2 3 x 3 + 2 3 ( 1 + x 2 ) 3 + C . 2. (a) Compute ˆ cos2 x + 6 sin2 x 2 d x . We expand the square and split in three integrals ˆ cos2 x + 6 sin2 x 2 d x = ˆ cos 2 2 x d x + 12 ˆ sin2 x cos2 x d x + 36 ˆ sin2 x d x . The first one yields ˆ cos 2 2 x d x = 1 2 ˆ cos4 x d x + 1 2 ˆ d x = 1 8 sin4 x + x 2 .
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