Learning Outcomes
- Integrate functions resulting in inverse trigonometric functions
Integrals that Result in Inverse Sine Functions
Let us begin with the three formulas. Along with these formulas, we use substitution to evaluate the integrals. We prove the formula for the inverse sine integral.
Integration Formulas Resulting in Inverse Trigonometric Functions
The following integration formulas yield inverse trigonometric functions:
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[latex]\displaystyle\int \frac{du}{\sqrt{{a}^{2}-{u}^{2}}}={ \sin }^{-1}\frac{u}{|a|}+C[/latex]
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[latex]\displaystyle\int \frac{du}{{a}^{2}+{u}^{2}}=\frac{1}{a}\phantom{\rule{0.05em}{0ex}}{ \tan }^{-1}\frac{u}{a}+C[/latex]
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[latex]\displaystyle\int \frac{du}{u\sqrt{{u}^{2}-{a}^{2}}}=\frac{1}{|a|}\phantom{\rule{0.05em}{0ex}}{ \sec }^{-1}\frac{|u|}{a}+C[/latex]
Proof
Let [latex]y={ \sin }^{-1}\dfrac{x}{a}.[/latex] Then [latex]a \sin y=x.[/latex] Now let’s use implicit differentiation. We obtain
For [latex]-\frac{\pi }{2}\le y\le \frac{\pi }{2}, \cos y\ge 0.[/latex] Thus, applying the Pythagorean identity [latex]{ \sin }^{2}y+{ \cos }^{2}y=1,[/latex] we have [latex]\cos y=\sqrt{1={ \sin }^{2}y}.[/latex] This gives
Then for [latex]\text{−}a\le x\le a,[/latex] we have
[latex]_\blacksquare[/latex]
Evaluating a Definite Integral Using Inverse Trigonometric Functions
Evaluate the definite integral [latex]{\displaystyle\int }_{0}^{\frac{1}{2}}\dfrac{dx}{\sqrt{1-{x}^{2}}}.[/latex]
Try It
Find the antiderivative of [latex]\displaystyle\int \frac{dx}{\sqrt{1-16{x}^{2}}}.[/latex]
Watch the following video to see the worked solution to the above Try It.
Example: Finding an Antiderivative Involving an Inverse Trigonometric Function
Evaluate the integral [latex]\displaystyle\int \frac{dx}{\sqrt{4-9{x}^{2}}}.[/latex]
Try It
Find the indefinite integral using an inverse trigonometric function and substitution for [latex]\displaystyle\int \frac{dx}{\sqrt{9-{x}^{2}}}.[/latex]
Watch the following video to see the worked solution to the above Try It.
Evaluating a Definite Integral
Evaluate the definite integral [latex]{\displaystyle\int }_{0}^{\sqrt{3}\text{/}2}\dfrac{du}{\sqrt{1-{u}^{2}}}.[/latex]
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.
Example: Finding an Antiderivative Involving the Inverse Tangent Function
Find an antiderivative of [latex]\displaystyle\int \frac{1}{1+4{x}^{2}}dx.[/latex]
Try It
Use substitution to find the antiderivative of [latex]\displaystyle\int \frac{dx}{25+4{x}^{2}}.[/latex]
Example: Applying the Integration Formulas
Find the antiderivative of [latex]\displaystyle\int \frac{1}{9+{x}^{2}}dx.[/latex]
Try It
Find the antiderivative of [latex]\displaystyle\int \frac{dx}{16+{x}^{2}}.[/latex]
Evaluating a Definite Integral
Evaluate the definite integral [latex]{\displaystyle\int }_{\sqrt{3}\text{/}3}^{\sqrt{3}}\dfrac{dx}{1+{x}^{2}}.[/latex]
Try It
Evaluate the definite integral [latex]{\displaystyle\int }_{0}^{2}\dfrac{dx}{4+{x}^{2}}.[/latex]
Watch the following video to see the worked solution to the above Try It.
Try It
Candela Citations
- 1.7 Integrals Resulting in Inverse Trigonometric Functions. Authored by: Ryan Melton. License: CC BY: Attribution
- Calculus Volume 1. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: https://openstax.org/details/books/calculus-volume-1. License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at https://openstax.org/books/calculus-volume-1/pages/1-introduction