Integrals Resulting in Inverse Trigonometric Functions

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{du}{\sqrt{a^2 - u^2}} = \arcsin \left(\frac{u}{a}\right) + C }

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{du}{a^2+u^2} =\dfrac{1}{a}\arctan \left(\dfrac{u}{a}\right) + C }

$\int {\frac {du}{u{\sqrt {u^{2}-a^{2}}}}}={\frac {1}{a}}\operatorname {arcsec} \left({\dfrac {|u|}{a}}\right)+C$

Example 1

Evaluate the integral

$\int {\dfrac {dx}{\sqrt {4-9x^{2}}}}.$

Solution

Substitute $u=3x$. Then 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 du=3dx} and we have

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\dfrac{dx}{\sqrt{4 - 9x^2}}=\dfrac{1}{3}\int\dfrac{du}{\sqrt{4 - u^2}}.}

Applying the formula with 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 a=2, } we obtain

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\dfrac{dx}{\sqrt{4 - 9x^2}}=\dfrac{1}{3}\int\dfrac{du}{\sqrt{4 - u^2}}=\dfrac{1}{3}\arcsin \left(\dfrac{u}{2}\right)+C=\dfrac{1}{3}\arcsin \left(\dfrac{3x}{2}\right)+C.}

Example 2

Evaluate 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}}\text{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}}\text{dx} = \int \frac{4}{\sqrt{16-x^2}}\text{dx} - \int \frac{x}{\sqrt{16-x^2}}\text{dx} }

The first integral is handled straightforward; the second integral is handled by substitution, with 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 u = 16-x^2} . We handle each separately.

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}{\sqrt{16-x^2}}\text{dx} = 4\arcsin\frac{x}{4} + C.}

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{x}{\sqrt{16-x^2}}\text{dx} } : Set 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 u = 16-x^2} , so 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 \text{du} = -2x\text{dx} } and 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 x\text{dx} = -\text{du} /2} . We have

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 \begin{align} \int\frac{x}{\sqrt{16-x^2}}\text{dx} = \int\frac{-\text{du} /2}{\sqrt{u}}\\ = -\frac12\int \frac{1}{\sqrt{u}}\text{du} \\ = - \sqrt{u} + C\\ = -\sqrt{16-x^2} + C.\end{align}}

Combining these together, we have

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}}\text{dx} = 4\arcsin\frac x4 + \sqrt{16-x^2}+C.}

Resources

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

• Integrating using Inverse Trigonometric Functions by patrickJMT

Licensing

Content obtained and/or adapted from:

• Integrals Resulting in Inverse Trigonometric Functions, LibreTexts: Mathematics under a CC BY-SA-NC license