HW13sol - = 3 cos θdθ Z 1(5-4 x-x 2 3 2 dx = Z 1(9-9 sin...

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M408C, Homework 13 — Solution Instructor: Guoliang Wu 1. Evaluate the following integral: Z tan 3 x sec xdx Solution: Let u = sec x , then du = sec x tan xdx and tan 2 x = sec 2 x - 1 = u 2 - 1. So Z tan 3 x sec xdx = Z tan 2 x sec x tan xdx = Z ( u 2 - 1) du = u 3 3 - u + C = sec 3 x 3 - sec x + C. 2. Evaluate the following integral: Z 1 (5 - 4 x - x 2 ) 3 / 2 dx Solution: Let’s first complete the square: 5 - 4 x - x 2 = 9 - (4+4 x + x 2 ) = 9 - ( x + 2) 2 . So we use the trig substitution: x + 2 = 3 sin θ and then dx
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Unformatted text preview: = 3 cos θdθ. Z 1 (5-4 x-x 2 ) 3 / 2 dx = Z 1 (9-9 sin 2 θ ) 3 / 2 (3 cos θ ) dθ = Z 3 cos θ (3 cos θ ) 3 dθ = 1 9 Z 1 cos 2 θ dθ = 1 9 Z sec 2 θdθ = 1 9 tan θ + C. Finally, since sin θ = x +2 3 , tan θ = x +2 √ 5-4 x-x 2 , and Z 1 (5-4 x-x 2 ) 3 / 2 dx = x + 2 9 √ 5-4 x-x 2 + C. 1...
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This note was uploaded on 12/03/2010 for the course M 408c taught by Professor Mcadam during the Fall '06 term at University of Texas.

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