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0910&0809T3S - MAT137Y 20092010 Winter Session...

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MAT137Y 2009–2010 Winter Session, Solutions to Term Test 3 1. Evaluate the following integrals. (8%) (i) Z dx x ( ln x ) 5 / 2 . Let u = ln x . Then du = 1 x dx , thereby giving us Z du u 5 / 2 = Z u - 5 / 2 du = - 2 3 u - 3 / 2 + C = - 2 3 ( ln x ) - 3 / 2 + C . (8%) (ii) Z 1 0 arctan xdx . We integrate by parts. Let u = arctan x , dv = dx . Then du = dx / ( 1 + x 2 ) and v = x . This yields, Z 1 0 arctan xdx = h x arctan x i 1 0 - Z 1 0 x 1 + x 2 dx = x arctan x - 1 2 ln ( 1 + x 2 ) 1 0 = arctan1 - 1 2 ln2 - 0 = π 4 - 1 2 ln2 . (10%) (iii) Z dx ( 2 + x 2 ) 2 . We apply a trig substitution: let x = 2tan θ . Then dx = 2sec 2 θ d θ . This gives Z 2sec 2 θ ( 2 + 2tan 2 θ ) 2 d θ = Z 2sec 2 θ 4sec 4 θ d θ = 2 4 Z cos 2 θ d θ . Using trig identities, 2 4 Z cos 2 θ d θ = 2 4 1 2 θ + 1 4 sin2 θ + C = 2 8 θ + 2 8 sin θ cos θ + C . We substitute back in terms of x . We draw a right triangle with x being the length of the opposite side and 2 being the length of the adjacent side. By the Pythagorean theorem the length of the hypoteneuse is x 2 + 2. Therefore sin θ = x x 2 + 2 and cos θ = 2 x 2 + 2 . Hence, Z dx ( 2 + x 2 ) 2 = 2 8 arctan x 2 + 2 8 x x 2 + 2 2 x 2 + 2 + C = 2 8 arctan x 2 + 1 4 x x 2 + 2 + C . (10%) (iv) Z 1 - x + 2 x 2 - x 3 x ( x 2 + 1 ) 2 dx . Let I be the integral above. We re-write the integral using partial fractions: I = Z A x + Bx + C x 2 + 1 + Dx + E ( x 2 + 1 ) 2 dx . 1
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We solve for the constants by grabbing a common denominator and matching coefficients with the original expression in the numerator: doing so gives us 1 - x + 2 x 2 - 2 x 3 = A ( x 2 + 1 ) 2 +( Bx + C ) x )( x 2 + 1 )+( Dx + E ) x = ( A + B ) x 4 + Cx 3 +( 2 A + B + D ) x 2 +( C + E ) x + A . Matching coefficients, we have A + B = 0 , C = - 1 , 2 A + B + D = 2 , C + E = - 1 , A = 1 . Solving for the remaining constants yields B = - 1, E = 0, D = 1. Therefore I = Z 1 x - x + 1 x 2 + 1 + x ( x 2 + 1 ) 2 dx = ln | x |- 1 2 ln ( x 2 + 1 ) - arctan x - 1 2 ( x 2 + 1 ) + C . 2. (10%) (i) Find the equation of the tangent line for the curve f ( x ) = ( tan x ) 1 / x at x = π 4 .
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