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# sol3-06 - MAT 137Y 2006-2007 Winter Session Solutions to...

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MAT 137Y 2006-2007 Winter Session, Solutions to Term Test 3 1. Evaluate the following integrals. (8%) (i) sec 3 x tan x dx . sec 3 x tan xdx = sec 2 x sec x tan xdx = 1 3 sec 3 x + C (by a simple substitution u = sec x ). (8%) (ii) 1 0 x · 3 x dx . Integrating by parts, let u = x , dv = 3 x dx . Then du = dx and v = 3 x ln3 . Hence 1 0 x · 3 x dx = x 3 x ln3 1 0 - 1 0 3 x ln3 dx = x 3 x ln3 - 3 x ( ln3 ) 2 1 0 = 3 ln3 - 3 ( ln3 ) 2 - 0 + 1 ( ln3 ) 2 = 3 ln3 - 2 ( ln3 ) 2 = 3ln3 - 2 ( ln3 ) 2 . (10%) (iii) dx x 4 x 2 - 1 . Let x = sec θ . Then dx = sec θ tan θ d θ , thus we have sec θ tan θ sec 4 θ sec 2 θ - 1 d θ = tan θ sec 3 θ tan θ d θ = d θ sec 3 θ = cos 3 θ d θ . To integrate cos 3 θ we take out one power of cos θ and apply a substitution: cos 3 θ d θ = ( 1 - sin 2 θ ) cos θ d θ = cos θ - sin 2 θ cos θ d θ = sin θ - 1 3 sin 3 θ + C . Substituting back, we can use triangles to see that if sec θ = x 1 , then sin θ = ( x 2 - 1 ) / x . Hence dx x 4 x 2 - 1 = x 2 - 1 x - 1 3 ( x 2 - 1 ) 3 / 2 x 3 + C . (10%) (iv) dx x 3 ( x + 1 ) 2 . We decompose the integrand using partial fractions: 1 x 3 ( x + 1 ) 2 = A x + B x 2 + C x 3 + D x + 1 + E ( x + 1 ) 2 Obtaining a common denominator on the right side (and ignoring the denominator) gives 1 = Ax 2 ( x + 1 ) 2 + Bx ( x + 1 ) 2 + C ( x + 1 ) 2 + Dx 3 ( x + 1 )+ Ex 3 = A ( x 4 + 2 x 3 + x 2 )+ B ( x 3 + 2 x 2 + x )+ C ( x 2 + 2 x + 1 )+ D ( x 4 + x 3 )+ Ex 3 = ( A + D ) x 4 +( 2 A + B + D + E ) x 3 +( A + 2 B + C ) x 2 +( B + 2 C ) x + C Matching coefficients gives us C = 1, B = - 2, A = 3, D = - 3, and E = - 1. Hence 3 x - 2 x 2 + 1 x 3 - 3 x + 1 - 1 ( x + 1 ) 2

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