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Practice Problems: Trig Substitution

Practice Problems: Trig SubstitutionWritten by Victoria 9, 2014 The following are solutions to the Trig Substitution Practice problems posted on November Use trig Substitution to show that 1 1 x2dx= sin 1x+CSolution:Letx= sin , thendx= cos : 1 1 x2dx= 1 1 sin2 cos d = cos cos d = d = +C= sin 1x+C 2. Use trig Substitution to show that 11+x2dx= tan 1x+CSolution:Letx= tan , thendx= sec2xdx: 11 +x2dx= 11 + tan2 sec2 d = sec2 sec2 d = d = +C= tan 1x+C 3. 1x2 x2+ 4dxSolution:Letx= 2 tan , thendx= 2 sec2 d : 1x2 x2+ 4dx= 14 tan2 4 tan2 + 4 2 sec2 d = 2 sec2 4 tan2 2 sec d = sec 4 tan2 d =14 cos sin2 d Now letw= sin , thendw= cos d :14 cos sin2 d =14 1w2dw= 14w+C= 14csc +CNext, we need to plug back inx. Originally we had the substitutionx= 2 tan , so tan = means our opposite side isx, our adjacent side is 2, and the hypotenuse is x2+ we have 1x2 x2+ 4dx= 14csc +C= x2+ 44x+C 14.

R 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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