Integrals Resulting in Inverse Trigonometric Functions

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

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

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

Example 1

Evaluate the integral

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

Solution

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

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

Applying the formula with $a=2,$ we obtain

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

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}}\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 $u=16-x^{2}$. We handle each separately.

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

$\int {\frac {x}{\sqrt {16-x^{2}}}}\ dx$: Set $u=16-x^{2}$, so $du=-2xdx<math>and<math>xdx=-du/2$. We have

${\begin{aligned}\int {\frac {x}{\sqrt {16-x^{2}}}}\ dx&amp;=\int {\frac {-du/2}{\sqrt {u}}}\\&amp;=-{\frac {1}{2}}\int {\frac {1}{\sqrt {u}}}\ du\\&amp;=-{\sqrt {u}}+C\\&amp;=-{\sqrt {16-x^{2}}}+C.\end{aligned}}$

Combining these together, we have

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

Resources

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

Integrating using Inverse Trigonometric Functions by patrickJMT