Exam 2 Solutions 2011

It says the right hand side of equation 3 is 25 2 1 1

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Unformatted text preview: c2 (θ) dθ 2 4x2 + 25 = 5 sec(θ) x= 4x2 + 25 dx = (5 sec(θ))(5/2) sec2 (θ) dθ = 25 2 sec3 (θ)dθ (3) The reduction formula comes in handy here. It says the right-hand side of equation (3) is 25 2 1 1 tan(θ) sec(θ) + sec(θ)dθ 2 2 25 25 tan(θ) sec(θ) + ln(|sec(θ) + tan(θ)|) + C = 4 4 so, written in terms of its original variable, x, the integral is √ √ 25 4x2 + 25 4x2 + 25 2x 25 2x + ln + 4 5 5 4 5 5 √ √ 2x 25 4x2 + 25 x 4x2 + 25 ln + + +C = 2 4 5 5 8. (10 points) Find +C dx . x(x + 3) Solution: Using the partial fractions approach, set 1 A B =+ x(x + 3) x x+3 Then 1 = A(x + 3) + Bx = (A + B )x + 3A 1 −1 =A =B 3 3 so 1 dx = x(x + 3) 3 d...
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This document was uploaded on 03/05/2014 for the course MAT 193 at CSU Dominguez Hills.

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