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Wize AP Calculus (AB) Textbook > Integration & Accumulation of Change

Antiderivatives of Inverse Trig Functions

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Antiderivatives (Indefinite Integrals) of Inverse Trig Functions

Let a>0a>0. Then we have the following indefinite integral rules:

∫1a2−x2dx=arcsin⁡(xa)+C\displaystyle \boxed{\int\frac{1}{\sqrt{a^2-x^2}}dx=\arcsin \left(\frac{x}{a}\right)+C}


∫1a2+x2dx=1aarctan⁡(xa)+C\displaystyle \boxed{\int\frac{1}{{a^2+x^2}}dx=\frac{1}{a}\arctan \left(\frac{x}{a}\right)+C}

∫1xx2−a2dx=1aarcsec(∣x∣a)+C\displaystyle \boxed{\int\frac{1}{x\sqrt{x^2-a^2}}dx=\frac{1}{a}\arcsec \left(\frac{|x|}{a}\right)+C}

Watch Out!
Different sources will state slightly different versions of the integral involving arcsecx\arcsec x.

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Example: Inverse Trig Indefinite Integrals

Evaluate the following indefinite integral

∫2−x2+9dx\displaystyle \int\frac{2}{\sqrt{-x^2+9}}dx

∫1a2−x2dx=arcsin⁡(xa)+C\displaystyle \int\frac{1}{\sqrt{a^2-x^2}}dx=\arcsin \left(\frac{x}{a}\right)+C

∫2−x2+9dx=∫2×19−x2dx\displaystyle \int\frac{2}{\sqrt{-x^2+9}}dx=\displaystyle \int2\times\frac{1}{\sqrt{9-x^2}}dx

=∫2×132−x2dx=\displaystyle \int2\times\frac{1}{\sqrt{3^2-x^2}}dx

=2arcsin⁡(x3)+C\displaystyle =\boxed{2\arcsin \left( \frac{x}{3}\right)+C}

Evaluate the following indefinite integral

∫14+2x2dx\displaystyle \int\frac{1}{4+2x^2}dx