- Calculate integrals that lead to inverse trigonometric function solutions
Integrals Resulting in Inverse Trigonometric Functions
The Main Idea
- Integration formulas yielding inverse trigonometric functions
- Domain restrictions for inverse trigonometric functions
- Connection between derivatives of inverse trig functions and these integrals
- Application of these formulas to various types of integrals
- Handling negative integrands in inverse trig integrals
Key Formulas
- [latex]\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}(\frac{u}{|a|}) + C[/latex]
- [latex]\int \frac{du}{a^2 + u^2} = \frac{1}{a}\tan^{-1}(\frac{u}{a}) + C[/latex]
- [latex]\int \frac{du}{u\sqrt{u^2 - a^2}} = \frac{1}{|a|}\sec^{-1}(\frac{|u|}{a}) + C[/latex]
Key Concepts
- Domain Restrictions:
- Inverse trig functions have restricted domains
- Solutions must respect these domain restrictions
- Recognizing Integrands:
- Identify integrands that match standard forms
- Use substitution to transform integrals into these forms
- Relationship to Derivatives:
- These integrals are related to derivatives of inverse trig functions
- Understanding this connection aids in application
- Handling Negative Integrands:
- Factor out [latex]-1[/latex] from negative integrands
- Use existing formulas rather than memorizing new ones
- Substitution Techniques:
- Often necessary to transform integrals into standard forms
- May involve adjusting constants or variables
Find the antiderivative of [latex]\displaystyle\int \frac{dx}{\sqrt{1-16{x}^{2}}}.[/latex]
Find the indefinite integral using an inverse trigonometric function and substitution for [latex]\displaystyle\int \frac{dx}{\sqrt{9-{x}^{2}}}.[/latex]
Use substitution to find the antiderivative of [latex]\displaystyle\int \frac{dx}{25+4{x}^{2}}.[/latex]
Find the antiderivative of [latex]\displaystyle\int \frac{dx}{16+{x}^{2}}.[/latex]
Evaluate the definite integral [latex]{\displaystyle\int }_{0}^{2}\dfrac{dx}{4+{x}^{2}}.[/latex]