sol3-04 - MAT 137Y 2004-2005 Solutions to Term Test 3 1...

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Unformatted text preview: MAT 137Y 2004-2005, Solutions to Term Test 3 1. Evaluate the following integrals. (10%) (i) Z sec 2 x tan x + 1 dx . Let u = tan x + 1. Then du = sec 2 x dx . Therefore, Z sec 2 x tan x + 1 dx = Z du u = ln | u | = ln | tan x + 1 | + C . (10%) (ii) Z p 9- x 2 dx . Let x = 3sin θ . Then dx = 3cos θ d θ . This gives Z 3cos θ · 3cos θ d θ = Z 9cos 2 θ d θ = 9 Z 1 + cos2 θ 2 d θ = 9 2 θ + 9 4 sin2 θ . Substituting back, we see that sin2 θ = 2sin θ cos θ = 2 · x 3 · √ 9- x 2 3 , so Z p 9- x 2 dx = 9 2 sin- 1 x 3 + 1 2 x p 9- x 2 + C . (10%) (iii) Z 2 x 2- x + 4 x ( x 2 + 4 ) dx . We decompose the rational function using partial fractions: 2 x 2- x + 4 x ( x 2 + 4 ) = A x + Bx + C x 2 + 4 = ⇒ A ( x 2 + 4 )+ Bx 2 + Cx = 2 x 2- x + 4 = ⇒ A + B = 2 C =- 1 4 A = 4 , so A = 1, B = 1, and C =- 1. Thus, Z 2 x 2- x + 4 x ( x 2 + 4 ) dx = Z 1 x + x- 1 x 2 + 4 dx = ln | x | + 1 2 ln ( x 2 + 4 )- 1 2 tan- 1 x 2 + C ....
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sol3-04 - MAT 137Y 2004-2005 Solutions to Term Test 3 1...

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