asst2soln

# asst2soln - MATH 138 Assignment 2-solution Due by 11 a.m...

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Unformatted text preview: MATH 138 Assignment 2-solution Due by: 11 a.m. , Friday, January 22. 1. Find the following indefinite integrals by making the suggested substitution. (a) Z e 2 x √ e x + 1 dx, x = ln( t 2 − 1) (b) Z dx 1 + √ x + 1 , x = t 2 − 1 (c) Z sin(ln x ) dx, x = e t . Solution: (a) Let x = ln( t 2 − 1) and dx = 2 t t 2 − 1 dt . Note that e x = t 2 − 1. Thus e 2 x = ( t 2 − 1) 2 and √ e x + 1 = √ t 2 = | t | . Without loss of generality, we can assume that t ≥ 0. We have Z e 2 x √ e x + 1 dx = Z ( t 2 − 1) 2 t · 2 t t 2 − 1 dt = Z 2( t 2 − 1) dt = 2 t 3 3 − 2 t + C = 2( e x + 1) 3 / 2 3 − 2 √ e x + 1 + C. (b) Let x = t 2 − 1 and dx = 2 tdt . Note that √ x + 1 = √ t 2 = | t | . Without loss of generality, we can assume that t ≥ 0. Then Z dx 1 + √ x + 1 = Z 2 t 1 + t dt = Z 2 − 2 1 + t dt = 2 t − 2 ln | 1 + t | + C = 2 √ x + 1 − 2 ln | 1 + √ x + 1 | + C. (c) Let x = e t and dx = e t dt . Note that ln x = ln( e t ) = t . We have Z sin(ln x ) dx = Z sin t · e t dt = e t sin t − e t cos t 2 + C (by Assignment 1, Question 3) = x sin(ln x ) − x cos(ln x ) 2 + C. 1 2. Find Z 2 1 √ x 2 − 1 x dx . Hint. Put x = sec θ . Also, if x runs covers the interval [1 , 2] what is the interval that θ covers? You have to know your trigonometry to proceed smoothly with this problem. Solution: Let x = sec θ (0 ≤ θ < π/ 2 or π ≤ θ < 3 π/ 2) and dx = tan θ sec θdθ . Then Z √ x 2 − 1 x dx = Z tan θ sec θ · tan θ sec θ dθ = Z tan 2 θ dθ = Z (sec 2 θ − 1) dθ = tan θ − θ + C....
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asst2soln - MATH 138 Assignment 2-solution Due by 11 a.m...

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