Integrals Resulting in Inverse Trigonometric Functions

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Example 1

Evaluate the integral

Solution

Substitute . Then and we have

Applying the formula with we obtain

Example 2

Evaluate Failed to parse (unknown function "\dx"): {\displaystyle \int \frac{4-x}{\sqrt{16-x^2}}\dx } .

Solution

This integral requires two different methods to evaluate it. We get to those methods by splitting up the integral:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{4-x}{\sqrt{16-x^2}}\ dx = \int \frac{4}{\sqrt{16-x^2}}\ dx - \int \frac{x}{\sqrt{16-x^2}}\dx } /p>

The first integral is handled straightforward; the second integral is handled by substitution, with . We handle each separately.

: Set , so . We have

Combining these together, we have

Resources

Integration into Inverse trigonometric functions using Substitution by The Organic Chemistry Tutor

Integrating using Inverse Trigonometric Functions by patrickJMT