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Unformatted text preview: Difference Equations Differential Equations to Section 6.6 Trigonometric Substitutions In the last section we saw that Z 1 1- x 2 dx = sin- 1 ( x ) + c. However, we arrived at this result as a consequence of our differentiation of the arc sine function, not as the outcome of the application of some systematic approach to the evalua- tion of integrals of this type. In this section we will explore how substitutions based on the arc sine, arc tangent, and arc secant functions provide a systematic method for evaluating integrals similar to this one. Sine substitutions To begin, consider evaluating Z 1 1- x 2 dx by using the substitution u = sin- 1 ( x ). The motivation for such a substitution stems from the fact that, for- 2 u 2 , u = sin- 1 ( x ) if and only if x = sin( u ). In the latter form, we see that dx = cos( u ) du and p 1- x 2 = q 1- sin 2 ( u ) = p cos 2 ( u ) = | cos( u ) | . (6.6.1) Since cos( u ) 0 when 2 u 2 , (6.6.1) becomes p 1- x 2 = cos( u ) . Thus Z 1 1- x 2 dx = Z cos( u ) cos( u ) du = Z du = u + c = sin- 1 ( x ) + c. Of course, there is nothing new in the result itself; it is the technique, which we may generalize to other integrals of a similar type, which is of interest. Specifically, for an integral with a factor of the form p a 2- x 2 or 1 a 2- x 2 , 1 Copyright c by Dan Sloughter 2000 2 Trigonometric Substitutions Section 6.6 where a > 0, the substitution x = a sin( u ) , 2 u 2 , (6.6.2) may prove to be useful because of the simplification p a 2- x 2 = q a 2- a 2 sin 2 ( x ) = q a 2 (1- sin 2 ( u )) = a p cos 2 ( u ) = a cos( u ) . (6.6.3) Although this substitution is equivalent to the substitution u = sin- 1 ( x a ) , we will see that it is more convenient to work with it in the form x = a sin( u ). Example To evaluate the integral Z 1 9- x 2 dx , we make the substitution x = 3 sin( u ) ,- 2 < u < 2 , dx = 3 cos( u ) du....
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- Spring '09