Practice Problems: Integration by Parts (Solutions)
vtkala@math.ucsb.edu November 25, 2014 The following are solutions to the Integration by Parts practice problems posted November 9. 1. R exsinxdx Solution: Let u= sinx, dv= exdx. Then du= cosxdxand v= ex. Then Z exsinxdx= exsinx Z excosxdx Now we need to use integration by parts on the second integral.
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web.math.ucsb.eduR x p 1 x4dx Solution: Z x p 1 x4dx= x 1 (x2)2dx Let u= x2, then du= 2xdx: Z x p 1 (x2)2dx= 1 2 Z 1 u2du Now let u= sin , then du= cos d : 1 2 Z p 1 u2du= 1 2 Z 1 sin2 cos d = 1 2 Z cos2 d = 1 4 Z (1+cos2 )d = 1 4 + 1 2 sin2 +C= 1 4 ( +sin cos )+C Plug back in u. Since u= sin , the opposite side will be u, the hypotenuse will be 1, and the
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