Integrals Resulting in Inverse Trigonometric Functions: Fresh Take

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Integrals Resulting in Inverse Trigonometric Functions: Fresh Take

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

$\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}(\frac{u}{|a|}) + C$
$\int \frac{du}{a^2 + u^2} = \frac{1}{a}\tan^{-1}(\frac{u}{a}) + C$
$\int \frac{du}{u\sqrt{u^2 - a^2}} = \frac{1}{|a|}\sec^{-1}(\frac{|u|}{a}) + C$

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 $-1$ 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 $\displaystyle\int \frac{dx}{\sqrt{1-16{x}^{2}}}.$

Substitute $u=4x$

$\frac{1}{4}\phantom{\rule{0.05em}{0ex}}{ \sin }^{-1}(4x)+C$

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “1.7 Integrals Resulting in Inverse Trigonometric Functions” here (opens in new window).

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

\sin^{- 1} (\frac{x}{3}) + C

${ \sin }^{-1}\left(\frac{x}{3}\right)+C$

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “1.7 Integrals Resulting in Inverse Trigonometric Functions” here (opens in new window).

Use substitution to find the antiderivative of $\displaystyle\int \frac{dx}{25+4{x}^{2}}.$

$\frac{1}{10}\phantom{\rule{0.05em}{0ex}}{ \tan }^{-1}\left(\frac{2x}{5}\right)+C$

Find the antiderivative of $\displaystyle\int \frac{dx}{16+{x}^{2}}.$

$\frac{1}{4}\phantom{\rule{0.05em}{0ex}}{ \tan }^{-1}(\frac{x}{4})+C$

Evaluate the definite integral ${\displaystyle\int }_{0}^{2}\dfrac{dx}{4+{x}^{2}}.$

$\frac{\pi }{8}$

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “1.7 Integrals Resulting in Inverse Trigonometric Functions” here (opens in new window).