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quiz2-sol - x back Notice that sec θ = 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 θ = x/ √ 7, we have tan θ = √ sec 2 θ-1 = p x 2 / 7-1 . Hence the integral is Z x √ x 2-7 dx = √ 7 tan θ + 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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