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# Calculus 2 Chapter 6.2 B MTH 162

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This 6 page Class Notes was uploaded by Jack Magann on Monday February 16, 2015. The Class Notes belongs to MTH 162 at University of Miami taught by Dr. Bibby in Spring2015. Since its upload, it has received 110 views. For similar materials see Calculus 2 in Mathematics (M) at University of Miami.

## Reviews for Calculus 2 Chapter 6.2 B

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Date Created: 02/16/15

Calculus 2 Chapter 62 B Trigonometric Substitution To get the idea of when trig substitution is used look at some past examples d 1f9967 sm 1 C xdx 1 2f fx9 x23dx u9 x2 du 2xdx 3f 2x9 x2dx lfudu 32u C 9 x2C 2 2 2 2 3 f 79sz This one can not be done with any of the methods previously used Problems like number 3 above are when trig substitution is used There are three instances where trig substitution is used 1 Integral involved Vaz u2 Set u asinH so that 9 sin 1 and du acos 9 d9 Place the known values on a right triangle so that Use SOHCAHTOA to relate to right triangle Hypotenuse a opposite side from angle u Pythagoras theorem gives adjacent side Vaz u2 We can then nd that Vaz u2 a cos 9 With these values we can substitute in the original equation Try the one above I xzdx W Letx3sin9 dx3c059 d9 9sin 1 V9 x2 Since cos 9 is adjacent over hypotenuse cos 9 2 Substitute in original equation I dex 3 sin923 cos 9d9 J9 29 d9 9 26d9 W 3 COS 9 s1n cos 99 sin29C 2 Integral involved Vaz u2 set u atan 9 so that 9 tan 1 and du asec2 9 d9 Place the known values on a right triangle Adjacent side a opposite side from angle u Pythagoras theorem gives Hypotenuse Vaz u2 We can nd that Vaz u2 a sec 9 3 Integral involved Vuz a2 set u asecH so that 9 2 sec 1 and du asecH tanH d9 Place the known values on a right triangle Adjacent side a Hypotenuse u Pythagoras theorem gives Opposite side Vuz a2 We can nd that Vuz a2 a tan 9 In all three cases you try to substitute the original equation with these new equations to make the derivative easier to nd Ex dx l f 3 Even though 1t 1s not to the 12 power Tr1g Substitutlon stlll applies 94x2 In this case u 2x 3 tanH 30 x 2 item 9 dx gsec2 9 d9 In this case it may help to draw the triangle and label the sides where Opposite side from angle 2x adjacent side 3 and hypotenuse V9 4x2 3 V94x2 Therefore W cos 9 sec 9 2 V9 4962 2 3sec9 3V9 4962 2 27sec39 Now substitute in these for the original equation f dx Bzfgsec26d6 1f d6 ifcosada 27 94x2 27 sec3 6 2 sec 6 18 1 39 9C 185m 3 f x9dx ux a3 Letx3sec9 dx3sec9tan9 If you draw a triangle Opposite side from angle sz 9 Adjacent side 3 Hypotenuse x Before substitution set the range of the integral to be in terms of 9 so you don t have to change the nal equation to be back in terms of x x3 33sec9 1sec9 90 x6 63sec9 2sec9 9 Substitute everything into the original equation f6Vx2 9dx 13 IStan0 3 x 035ec0 3 sec 9 tan 9d9 f0 3 tan2 9 d9 Now use the Pythagorean identity tan2 9 2 sec2 9 1 f 3tan29d9 3f sec29 1d9 3tan9 6 3 g0 3J n 3f Vx2 4x5 x 1 dx For this multiple methods from previous chapters need to be used Start by completing the square in the bottom x2 4x5 x 1 Vx2 4x5 dx x2 4x41x 221 W Use the W substitution method wx 2 xw2 x 1 fwx 221 dx dx dw Now substitute f W1 dW Vw21 Now split this up into two integrals to solve separately W1 Vw21 f dwf W 1 mdwfmdw The rst can be solved using u substitution f f MI 1 2 2 l d u Wz1dw u w 1 du dew deW Zf u The second is solve using trig substitution 2 W w 1tan9 dwsec29 secH w11Vw21 In triangle Opposite w adjacent 1 and hypotenuse w2 1 2 viz1 fscz9 fseCH d9 1nsec9 tanHI 1nW2 1 WI The nal answer should then be de w21ln w21wC Vx2 4x5 x 1 Vx2 4x5 OI39 dx x 2211n x 221x 2C w21 x 1 f dxxx2 4x5lnx2 4x5x 2C Vx2 4x5 All these notes are from a Calculus 2 class run by Dr Patrick Dibby at the University of Miami A majority of the problems used by the teacher and copied in these notes are taken from the textbook Essential Calculus Second edition By James Stewart Citation Stewart James Essential Calculus 2nd ed Np Cangage Learning 2013 Print

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