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 \(\displaystyle \int \frac{4-x}{\sqrt{16-x^2}}\ dx\).

Solution

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

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

The first integral is handled using a straightforward application of Theorem \(\PageIndex{2}\); the second integral is handled by substitution, with \(u = 16-x^2\). We handle each separately.

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

\(\displaystyle \int\frac{x}{\sqrt{16-x^2}}\ dx\): Set \(u = 16-x^2\), so \(du = -2xdx\) and \(xdx = -du/2\). We have

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

Combining these together, we have

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

Resources

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

Integrating using Inverse Trigonometric Functions by patrickJMT