quiz2-sol - x back. Notice that sec = x/ 7, we have tan =...

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Quiz 2 – Math 2153, Calculus II – Sept. 2, 2011 1. Evaluate the integral Z x x 2 - 7 dx Make sure your solution does NOT contain any inverse trigonometric function. Solution 1 Use trigonometric substitution. Set x = 7 sec θ , where θ [0 , π 2 ) [ π, 3 π 2 ), then dx = 7 sec θ tan θ dθ . The integral can be written as Z x x 2 - 7 dx = Z 7 sec θ 7 sec 2 θ - 7 7 sec θ tan θ dθ = Z 7 sec θ 7 tan θ 7 sec θ tan θ dθ = Z 7 sec 2 θ dθ = 7 tan θ + C Now we need to substitute
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Unformatted text preview: x back. Notice that sec = x/ 7, we have tan = sec 2 -1 = p x 2 / 7-1 . Hence the integral is Z x x 2-7 dx = 7 tan + C = 7 p x 2 / 7-1 + C = x 2-7 + C Solution 2 Use substitution. Set u = x 2-7, then du = 2 x dx The integral becomes Z x x 2-7 dx = 1 2 u du = u + C = x 2-7 + C...
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This note was uploaded on 09/22/2011 for the course MATH 2153 taught by Professor Staff during the Fall '08 term at Oklahoma State.

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