610
Chapter 5 | Integration
Solution Substitute u =3 x . Then du =3 dx and we have
⌠ ⌡
⌠ ⌡
dx 4−9 x 2
du 4− u 2
= 1 3
.
Applying the formula with a =2, we obtain ⌠ ⌡ dx 4−9 x 2
⌠ ⌡
du 4− u 2
= 1 3
−1 ⎛ −1 ⎛
⎞ ⎠ + C
⎝ u 2
= 1 3 sin = 1 3 sin
⎞ ⎠ + C .
⎝ 3 x 2
Find the indefinite integral using an inverse trigonometric function and substitution for ⌠ ⌡ dx 9− x 2 .
5.41
Example 5.51 Evaluating a Definite Integral Evaluate the definite integral ⌠ ⌡ 0 3/2 du 1− u 2 .
Solution The format of the problem matches the inverse sine formula. Thus, ⌠ ⌡ 0 3/2 du 1− u 2 = sin −1 u | 0 3/2 = ⎡ ⎣ sin −1 ⎛ ⎝ 3 2 ⎞ ⎠
⎤ ⎦ −
⎡ ⎣ sin −1 (0) ⎤ ⎦
= π
3 .
Integrals Resulting in Other Inverse Trigonometric Functions There are six inverse trigonometric functions. However, only three integration formulas are noted in the rule on integration formulas resulting in inverse trigonometric functions because the remaining three are negative versions of the ones we use. The only difference is whether the integrand is positive or negative. Rather than memorizing three more formulas, if the integrand is negative, simply factor out −1 and evaluate the integral using one of the formulas already provided. To close this section, we examine one more formula: the integral resulting in the inverse tangent function.
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