Chapter 5 | Integration
609
1 a cos y = 1
a 1−sin 2 y
= 1
a 2 − a 2 sin 2 y
= 1
.
a 2 − x 2
Then for − a ≤ x ≤ a , and generalizing to u , we have ⌠ ⌡ 1 a 2 − u 2
du = sin −1 ⎛
⎞ ⎠ + C .
⎝ u a
□
Example 5.49 Evaluating a Definite Integral Using Inverse Trigonometric Functions
1 2
Evaluate the definite integral ⌠ ⌡ 0
dx 1− x 2
.
Solution We can go directly to the formula for the antiderivative in the rule on integration formulas resulting in inverse trigonometric functions, and then evaluate the definite integral. We have
1 2
1 2
⌠ ⌡ 0
= sin −1 x | = sin −1 1
dx 1− x 2
0
−1 0
2 −sin
= π 6 = π 6
−0
.
Find the antiderivative of ⌠ ⌡
5.40
dx 1−16 x 2 .
Example 5.50 Finding an Antiderivative Involving an Inverse Trigonometric Function Evaluate the integral ⌠ ⌡ dx 4−9 x 2 .
Made with FlippingBook - professional solution for displaying marketing and sales documents online