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. Assume \(a > 0\):

\[\int \frac{d u}{\sqrt{a^{2} - u^{2}}} = \text{sin}^{- 1} \frac{u}{a} + C\]
\[\int \frac{d u}{a^{2} + u^{2}} = \frac{1}{a} \text{tan}^{−1} \frac{u}{a} + C\]
\[\int \frac{d u}{u \sqrt{u^{2} - a^{2}}} = \frac{1}{a} \text{sec}^{−1} \frac{\left|\right. u \left|\right.}{a} + C\]

Proof

Let \(y = \text{sin}^{−1} \frac{x}{a} .\) Then \(a \text{sin} y = x .\) Now let’s use implicit differentiation. We obtain

\[\begin{aligned} \frac{d}{d x} \left(\right. a \text{sin} y \left.\right) & = & \frac{d}{d x} \left(\right. x \left.\right) \\ \\ a \text{cos} y \frac{d y}{d x} & = & 1 \\ \frac{d y}{d x} & = & \frac{1}{a \text{cos} y} . \end{aligned}\]

For \(- \frac{\pi}{2} \leq y \leq \frac{\pi}{2} , \text{cos} y \geq 0 .\) Thus, applying the Pythagorean identity \(\text{sin}^{2} y + \text{cos}^{2} y = 1 ,\) we have \(\text{cos} y = \sqrt{1 – \text{sin}^{2} y} .\) This gives

\[\begin{aligned} \frac{1}{a \text{cos} y} & = \frac{1}{a \sqrt{1 - \text{sin}^{2} y}} \\ \\ & = \frac{1}{\sqrt{a^{2} - a^{2} \text{sin}^{2} y}} \\ & = \frac{1}{\sqrt{a^{2} - x^{2}}} . \end{aligned}\]

Then for \(− a \leq x \leq a ,\) and generalizing to u, we have

\[\int \frac{1}{\sqrt{a^{2} - u^{2}}} d u = \text{sin}^{−1} \left(\right. \frac{u}{a} \left.\right) + C .\]

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Example 5.49

Evaluating a Definite Integral Using Inverse Trigonometric Functions

Evaluate the definite integral \(\int_{0}^{\frac{1}{2}} \frac{d x}{\sqrt{1 - x^{2}}} .\)

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

\[\begin{aligned} \\ \\ \int_{0}^{\frac{1}{2}} \frac{d x}{\sqrt{1 - x^{2}}} & = \text{sin}^{−1} x \left|\right._{0}^{\frac{1}{2}} \\ & = \text{sin}^{−1} \frac{1}{2} - \text{sin}^{−1} 0 \\ & = \frac{\pi}{6} - 0 \\ & = \frac{\pi}{6} . \end{aligned}\]

Checkpoint 5.40

Evaluate the integral \(\int \frac{d x}{\sqrt{1 - 16 x^{2}}} .\)

Example 5.50

Finding an Antiderivative Involving an Inverse Trigonometric Function

Evaluate the integral \(\int \frac{d x}{\sqrt{4 - 9 x^{2}}} .\)

Solution

Substitute \(u = 3 x .\) Then \(d u = 3 d x\) and we have

\[\int \frac{d x}{\sqrt{4 - 9 x^{2}}} = \frac{1}{3} \int \frac{d u}{\sqrt{4 - u^{2}}} .\]

Applying the formula with \(a = 2 ,\) we obtain

\[\begin{aligned} \int \frac{d x}{\sqrt{4 - 9 x^{2}}} & = \frac{1}{3} \int \frac{d u}{\sqrt{4 - u^{2}}} \\ \\ & = \frac{1}{3} \text{sin}^{−1} \left(\right. \frac{u}{2} \left.\right) + C \\ & = \frac{1}{3} \text{sin}^{−1} \left(\right. \frac{3 x}{2} \left.\right) + C . \end{aligned}\]

Checkpoint 5.41

Find the indefinite integral using an inverse trigonometric function and substitution for \(\int \frac{d x}{\sqrt{9 - x^{2}}} .\)

Example 5.51

Evaluating a Definite Integral

Evaluate the definite integral \(\int_{0}^{\sqrt{3} / 2} \frac{d u}{\sqrt{1 - u^{2}}} .\)

Solution

The format of the problem matches the inverse sine formula. Thus,

\[\begin{aligned} \\ \\ \int_{0}^{\sqrt{3} / 2} \frac{d u}{\sqrt{1 - u^{2}}} & = \text{sin}^{−1} u \left|\right._{0}^{\sqrt{3} / 2} \\ & = \left[\right. \text{sin}^{−1} \left(\right. \frac{\sqrt{3}}{2} \left.\right) \left]\right. - \left[\right. \text{sin}^{−1} \left(\right. 0 \left.\right) \left]\right. \\ & = \frac{\pi}{3} . \end{aligned}\]

This lesson is part of:

Integration

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