Midterm 2 Solutions - Math 31B Midterm 2 Solutions Ian...

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Math 31B, Midterm 2 Solutions Ian Coley December 2, 2014 Solution to 1. (a) We need to find Z π/ 2 0 cos 3 x dx. We should first use the substitution 1 - sin 2 x = cos 2 x . This gives Z π/ 2 0 (1 - sin 2 x ) cos x dx. Let u = sin x , so du = cos x dx . This also gives us the new bounds sin 0 = 0 and sin( π/ 2) = 1. Hence Z π/ 2 0 cos 3 x dx = Z 1 0 1 - u 2 du = u - u 3 3 1 0 = 1 - 1 3 - (0 - 0) = 2 3 . (b) This question is best done with partial fractions. It turns out to be possible if you don’t split x 2 - 4 = ( x - 2)( x + 2), but that’s sloppy so we won’t do it. We want x - 4 ( x - 1)( x 2 - 4) = A x - 4 + B x - 2 + C x + 2 . Multiplying through by the denominator, we get x - 4 = A ( x - 2)( x + 2) + B ( x - 1)( x + 2) + C ( x - 1)( x - 2) . We can solve this by plugging in various values for x . Starting with x = 1, this zeroes out the B and C term, so 1 - 4 = A ( - 1)(3) = A = 1 . For x = 2, this zeroes out A and C , so we get 2 - 4 = B (2 - 1)(2 + 2) = B = - 1 / 2 . For x = - 2, this zeroes out A and B , so we get - 2 - 4 = C ( - 2 - 1)( - 2 - 2) = C = - 1 / 2 . 1
Overall, we then have x - 4 ( x - 1)( x 2 - 4) = 1 x - 4 - 1 2 · 1 x - 2 - 1 2 · 1 x + 2 . Therefore Z x - 4 ( x - 1)( x 2 - 4) dx = Z 1 x - 4 dx - 1 2 Z 1 x - 2 dx - 1 2 Z 1 x + 2 dx = ln | x - 4 | - 1 2 ln | x - 2 | - 1 2 ln | x + 2 | + C. Solution to 2. Let I m = R π/ 2 0 sin m x dx . Then we’d like to show I m = m - 1 m I m - 2 for m 2 . To begin with, we should take the above integral and do a clever integration by parts. Let u = sin m - 1 x and dv = sin x dx . This gives du = ( m - 1) sin m - 2 x cos x dx

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