Exam 1 Sol'n

# Exam 1 Sol'n - C HECKMAN MAT 266 Solutions 12:15 Class Test...

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Unformatted text preview: C. HECKMAN MAT 266 Solutions, 12:15 Class Test 1 Solutions to the 1:40 class’s version of the test appear five pages later. (1) [15 points] Do a trigonometric substitution on the following integral. That is, make a substitution which gets rid of the radical signs and all the x ’s. You may stop when you have gotten rid of these things; you do not need to evaluate the resulting integral. Z e 3 x sin ( √ x 2 + 16 + 4 x ) 1 + ln ( x- √ x 2 + 16 ) dx Solution: Because of the presence of p x 2 + 16 = p x 2 + 4 2 , the proper trigono- metric substitution is x = 4 tan θ . This means dx = 4 sec 2 θ dθ and p x 2 + 16 = 4 sec θ . Substituting these into the integral produces Z e 3 · 4 tan θ sin ( 4 sec θ + 4 · 4 tan θ ) 1 + ln ( 4 tan θ- 4 sec θ ) · 4 sec 2 θ dθ. Grading: +4 points for choosing the trigonometric substitution, +4 points for dx and the square root, +4 points for substituting into the function, +3 points for replacing dx . 1 (2) Find the following integrals: (a) [10 points] Z x 2 e 3 x dx Solution: Integration by parts, twice, then a u-substitution: Z x 2 e 3 x dx = x 2 · e 3 x 3- Z 2 x · e 3 x 3 dx u = x 2 v = e 3 x u = 2 x v = e 3 x 3 = x 2 e 3 x 3- 2 3 Z xe 3 x dx = x 2 e 3 x 3- 2 3 x · e 3 x 3- Z 1 · e 3 x 3 dx u = x v = e 3 x u = 1 v = e 3 x 3 = x 2 e 3 x 3- 2 3 xe 3 x 3- e 3 x 9 + C. Grading: +4 points for finding the first integration by parts, +2 points for doing it, +2 + 2 points for finding and doing the next integration by parts. (b) [10 points] Z 3 x + 1 x 2 + 81 dx Solution: This is already a partial fraction, since the denominator has no real roots. You had to split this fraction into two pieces, do a u-substitution on the first one, and use the arc tangent formula on the second one. Z 3 x + 1 x 2 + 81 dx = 3 Z x x 2 + 81 dx + Z 1 x 2 + 81 dx = 3 · 1 2 ln( x 2 + 81) + 1 9 tan- 1 x 9 + C. Grading: +4 points for splitting up the fraction, +3 points for each piece. = ⇒ 2 (c) [10 points] Z sin 6 θ cos 5 θ dθ Solution: Since the power of cos θ is odd, you should factor it out, convert the rest of the function into sin θ ’s using the identity sin 2 θ + cos 2 θ = 1 , and make the substitution u = sin θ : Z sin 6 θ cos 5 θ dθ = Z sin 6 θ cos 4 θ · cos θ dθ = Z sin 6 θ (1- sin 2 θ ) 2 · cos θ dθ = Z u 6 (1- u 2 ) 2 du u = sin θ du dθ = cos θ du = cos θ dθ = Z u 6 (1- 2 u 2 + u 4 ) du = Z u 6- 2 u 8 + u 10 du = u 7 7- 2 9 u 9 + u 11 11 + C = sin 7 θ 7- 2 9 sin 9 θ + sin 11 θ 11 + C. Grading: +3 points for the idea, +2 points for converting to sin θ ’s, +2 points for multiplying out the polynomial, +3 points for finishing up....
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## This note was uploaded on 10/05/2008 for the course MAT 266 taught by Professor Farisodish during the Spring '08 term at ASU.

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Exam 1 Sol'n - C HECKMAN MAT 266 Solutions 12:15 Class Test...

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