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

Solution

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

$\int {\frac {4-x}{\sqrt {16-x^{2}}}}{\text{dx}}=\int {\frac {4}{\sqrt {16-x^{2}}}}{\text{dx}}-\int {\frac {x}{\sqrt {16-x^{2}}}}{\text{dx}}$ /p>

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

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

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

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