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Solutions Calculus II - Spring 2013 - Written homework 6 Z 1. Integrate (give your answer in terms of t): √ 1 dt 9 − t2 We should rewrite this one prior to substituting: Z Z 1 1 √ q dt = 2 9−t 9 1− Z t2 dt = 9 1 q 3 1− t 2 3 dt = sin θ, so dt = 3 cos θ dθ, and, for back-substitution purposes, we have θ = arcsin 3t : Z Z Z 1 3 cos θ dθ cos θ p √ dθ = 1 dθ = 3 cos2 θ 1 − sin2 θ t = θ + C = arcsin +C 3 Z 3x √ 2. Integrate (give your answer in terms of x): dx x2 − 4 Now, let t 3 Notice that this one could easily be done with a regular substitution! But, if you didn’t catch that, then you could have proceeded with a trigonometric substitution, as follows. First, it must be rewritten a bit: Z Z Z Z 3x x x x 3 √ q q dx = 3 dx = 3 q dx = dx 2 2 2 x2 x2 − 4 x x 4 4 −1 2 − 1 − 1 2 2 Now it is clear that the substitution should be x2 = sec θ, whence x = 2 sec θ, dx = 2 sec θ tan θ dθ, and for back-substitution purposes, we note that θ = arcsec x2 . Rewrite the integral to get: 3 2 Z √ 2 sec θ 2 sec θ tan θ dθ = 6 sec2 θ − 1 Z sec2 θ tan θ √ dθ = 6 tan2 θ = 6 tan θ + C = 6 tan arcsec Z sec2 θ dθ x +C 2 Again, we use a triangle to recover tan arcsec x2 : √ x2 − 4. Thus, the final answer becomes 3 x2 − 4 + C. Z x2 3. Integrate (give your answer in terms of x): dx 3/2 (1 + x2 ) From this we calculate tan arcsec x2 = 1 2 √ Using trigonometric substitution, let x = tan θ. Recall for our later back substitution that this means θ = arctan x. Then dx = sec2 θ dθ, and the integral becomes: Z Z Z Z tan2 θ sec2 θ tan2 θ sec2 θ tan2 θ sec2 θ tan2 θ dθ = dθ = dθ = dθ 3/2 3/2 sec3 θ sec θ (sec2 θ) 1 + tan2 θ 1 Z = Z = sin2 θ dθ = cos θ Z sin2 θ cos2 θ 1 cos θ Z dθ = 1 − cos2 θ dθ = cos θ Z cos θ dθ sin2 θ 1 − cos θ dθ = cos θ Z sec θ − cos θ dθ = ln | sec θ + tan θ| − sin θ + C = ln |sec arctan x + x| − sin arctan x + C To recover an elementary value for x, sketch a right triangle with an angle θ whose tangent is x, find the missing third side of the triangle, and use this to find the other values: √ x From this we get sin arctan x = √1+x , and sec arctan x = x2 + 1, so the solution to the integral is 2 p x + C. ln x2 + 1 + x − √ x2 + 1 Z 7 dx 4. Integrate, using partial fraction decomposition: (x − 1)(2x + 4) We set up the sum as follows: A B 7 = + , whence (x − 1)(2x + 4) x − 1 2x + 4 7 2Ax + 4A + Bx − B = (x − 1)(2x + 4) (x − 1)(2x + 4) This gives us two equations to solve, 4A − B = 7, and 2A + B = 0. Solve these using elementary algebra to get B = − 37 and A = 76 ; this yields the integral Z 7 1 1 7 7 x − 1 − = (ln |x − 1| − ln |x + 2|) + C = ln +C 6 x−1 x+2 6 6 x + 2 2