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Chapter 5 | Integration
5.7 | Integrals Resulting in Inverse Trigonometric Functions Learning Objectives 5.7.1 Integrate functions resulting in inverse trigonometric functions In this section we focus on integrals that result in inverse trigonometric functions. We have worked with these functions before. Recall from Functions and Graphs that trigonometric functions are not one-to-one unless the domains are restricted. When working with inverses of trigonometric functions, we always need to be careful to take these restrictions into account. Also in Derivatives , we developed formulas for derivatives of inverse trigonometric functions. The formulas developed there give rise directly to integration formulas involving inverse trigonometric functions. Integrals that Result in Inverse Sine Functions Let us begin this last section of the chapter with the three formulas. Along with these formulas, we use substitution to evaluate the integrals. We prove the formula for the inverse sine integral.
Rule: Integration Formulas Resulting in Inverse Trigonometric Functions The following integration formulas yield inverse trigonometric functions: 1.
⌠ ⌡
(5.23)
du a 2 − u 2
= sin −1 u
C
| a | +
2.
⌠ ⌡
(5.24)
du a 2 + u 2
−1 u
= 1 a tan
a + C
3.
⌠ ⌡
(5.25)
du u u 2 − a 2
−1 u
= 1 | a | sec
C
| a | +
Proof Let y = sin −1 x
a . Then a sin y = x . Now let’s use implicit differentiation. We obtain d dx ⎛ ⎝ a sin y ⎞ ⎠ = d dx ( x ) a cos y dy dx = 1 dy dx = 1 a cos y .
For − π
y ≤ π
2 , cos y ≥0. Thus, applying the Pythagorean identity sin 2 y +cos 2 y =1,
we have
2 ≤
cos y = 1 – sin 2 y . This gives
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