Integrals Resulting in Inverse Trigonometric Functions

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

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

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

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

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

Resources

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

Integrating using Inverse Trigonometric Functions by patrickJMT