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

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 \[ ∫\dfrac{dx}{\sqrt{4−9x^2}}.\nonumber\]}

Solution

Substitute Failed to parse (syntax error): {\displaystyle \( u=3x\)} . Then Failed to parse (syntax error): {\displaystyle \( du=3\,dx\)} and we have

Failed to parse (syntax error): {\displaystyle \[ ∫\dfrac{dx}{\sqrt{4−9x^2}}=\dfrac{1}{3}∫\dfrac{du}{\sqrt{4−u^2}}.\nonumber\]}

Applying the formula with Failed to parse (syntax error): {\displaystyle \( a=2,\)} we obtain

Failed to parse (syntax error): {\displaystyle \[ ∫\dfrac{dx}{\sqrt{4−9x^2}}=\dfrac{1}{3}∫\dfrac{du}{\sqrt{4−u^2}}=\dfrac{1}{3}\arcsin \left(\dfrac{u}{2}\right)+C=\dfrac{1}{3}\arcsin \left(\dfrac{3x}{2}\right)+C.\nonumber\]}

Resources

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

Integrating using Inverse Trigonometric Functions by patrickJMT