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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.
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.
The following integration formulas yield inverse trigonometric functions:
Let $y={\text{sin}}^{\mathrm{-1}}\frac{x}{a}.$ Then $a\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}y=x.$ Now let’s use implicit differentiation. We obtain
For $-\frac{\pi}{2}\le y\le \frac{\pi}{2},\text{cos}\phantom{\rule{0.1em}{0ex}}y\ge 0.$ Thus, applying the Pythagorean identity ${\text{sin}}^{2}y+{\text{cos}}^{2}y=1,$ we have $\text{cos}\phantom{\rule{0.1em}{0ex}}y=\sqrt{1={\text{sin}}^{2}y}.$ This gives
Then for $\text{\u2212}a\le x\le a,$ we have
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Evaluate the definite integral ${\int}_{0}^{1}\frac{dx}{\sqrt{1-{x}^{2}}}.$
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
Find the antiderivative of $\int \frac{dx}{\sqrt{1-16{x}^{2}}}}.$
$\frac{1}{4}\phantom{\rule{0.05em}{0ex}}{\text{sin}}^{\mathrm{-1}}\left(4x\right)+C$
Evaluate the integral $\int \frac{dx}{\sqrt{4-9{x}^{2}}}}.$
Substitute $u=3x.$ Then $du=3dx$ and we have
Applying the formula with $a=2,$ we obtain
Find the indefinite integral using an inverse trigonometric function and substitution for $\int \frac{dx}{\sqrt{9-{x}^{2}}}}.$
${\text{sin}}^{\mathrm{-1}}\left(\frac{x}{3}\right)+C$
Evaluate the definite integral ${\int}_{0}^{\sqrt{3}\text{/}2}\frac{du}{\sqrt{1-{u}^{2}}}.$
The format of the problem matches the inverse sine formula. Thus,
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.
Find an antiderivative of $\int \frac{1}{1+4{x}^{2}}dx.$
Comparing this problem with the formulas stated in the rule on integration formulas resulting in inverse trigonometric functions, the integrand looks similar to the formula for ${\text{tan}}^{\mathrm{-1}}u+C.$ So we use substitution, letting $u=2x,$ then $du=2dx$ and $1\text{/}2du=dx.$ Then, we have
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