105-4-TrigSubsPartialFracsNumInt

# 105-4-TrigSubsPartialFracsNumInt - 4 Trigonometric...

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Unformatted text preview: 4 Trigonometric Substitutions, Partial Fractions, and Numerical Integration 4.1 Trigonometric Substitutions Introduction We do not yet know how to handle integrals involving variations of the functions 1- x 2 , 1 + x 2 , x 2- 1 . For instance, we cannot yet do integrals like Z 1- x 2 dx, Z 1 4 + x 2 , Z x 2 (2 x 2- 3) 3 / 2 . The basic observation that we need for solving these is that q 1- sin 2 ( ) * = cos( ) , q 1 + tan 2 ( ) * = sec( ) , p sec 2 ( )- 1 * = tan( ) . Ive put a * over these equalities because theyre not quite true: the right hand sides should be | cos( ) | , | sec( ) | , and | tan( ) | , since x 2 = | x | , not x . But Im going to ignore this for now, and go into the finer details later. Here is why these equations help: to do for instance R 1- x 2 dx , we can use the substitution x = sin( ) dx = cos( ) d to get Z 1- x 2 dx = Z q 1- sin 2 ( ) (cos( ) d ) = Z cos 2 ( ) d, and this is a trigonometric integral that we can do. Note that this substitution is a bit different from the ones were used to: instead of something like u = x 2 when we see x 2 in the integrand, were writing x = sin( ), even though we dont see a sin( ) in the integrand. Similarly, we can use the substitutions x = tan( ), dx = sec 2 ( ) d to do Z 1 + x 2 dx = Z q 1 + tan 2 ( ) (sec 2 ( ) d ) = Z sec 3 ( ) d, an integral we saw in the last chapter. And we can use x = sec( ), dx = sec( )tan( ) d for Z x 2- 1 dx = Z p sec 2 ( )- 1 (sec( )tan( ) d ) = Z tan 2 ( )sec( ) d, which we can finish using tan 2 ( ) = 1- sec 2 ( ). 1 Different forms of 1- x 2 , 1 + x 2 , x 2- 1 . There are a number of different forms of these root functions for which we can use these substitutions. I will illustrate these here without doing the integrals yet. For instance, for 9- x 2 we couldnt use x = sin( ) itself, but we can modify it to x = 3sin( ): 9- x 2 x =3 sin( ) = q 3 2- 3 2 sin 2 ( ) = 3 q 1- sin 2 ( ) = 3cos( ) . Here are some examples of modified substitutions that work, with increasing difficulty: 1 + 4 x 2 x = 1 2 tan( ) = q 1 + 2 2 (1 / 2) 2 tan 2 ( ) = q 1 + tan 2 ( ) = sec( ) , p ( x + 1) 2- 25 x =5 sec( )- 1 = p (5sec( )) 2- 5 2 = 5 p sec 2 ( )- 1 = 5tan( ) , 2- 3 x 2 x = 2 3 sin( ) = q 2- 3 ( 2 / 3) 2 sin 2 ( ) = q 2- 2sin 2 ( ) = 2cos( ) . Examples of integrals Z 1 1- x 2 dx We actually already know an antiderivative for the integrand, but its a good example, so lets pretend we dont know that antiderivative. Since we see a 1- x 2 , we should use the substitution x = sin( ), dx = cos( ) d , so Z 1 1- x 2 dx = Z 1 p 1- sin 2 ( ) cos( ) d = Z cos( ) cos( ) d = Z d = + C, which is not the answer yet, because we have to return to...
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## This note was uploaded on 01/14/2012 for the course MATH 105 taught by Professor Malabikapramanik during the Fall '10 term at The University of British Columbia.

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105-4-TrigSubsPartialFracsNumInt - 4 Trigonometric...

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