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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 π 2 ≤ u ≤ π 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 π 2 ≤ u ≤ π 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 ) , π 2 ≤ u ≤ π 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 ) , π 2 < u < π 2 , dx = 3 cos( u ) du....
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 Spring '09
 Johnson
 Algebra

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