Six Inverse Trig Functions

Unit 6: Antiderivatives Six Inverse Trig Functions

Base Case: Six Inverse Trig Functions

Base Case	Definite Integral	Indefinite Integral
$f ‘ (x) = \frac{1}{\sqrt{1 - x^{2}}}$	$\int_{a}^{b} \frac{1}{\sqrt{1 - x^{2}}} dx = {\sin^{- 1} (x) \|}_{x = a}^{x = b}$	$\int \frac{1}{\sqrt{1 - x^{2}}} dx = \sin^{- 1} (x) + C$
$f ‘ (x) = \frac{- 1}{\sqrt{1 - x^{2}}}$	$\int_{a}^{b} \frac{- 1}{\sqrt{1 - x^{2}}} dx = {\cos^{- 1} (x) \|}_{x = a}^{x = b}$	$\int \frac{- 1}{\sqrt{1 - x^{2}}} dx = \cos^{- 1} (x) + C$
$f ‘ (x) = \frac{1}{1 + x^{2}}$	$\int_{a}^{b} \frac{1}{1 + x^{2}} dx = {\tan^{- 1} (x) \|}_{x = a}^{x = b}$	$\int \frac{1}{1 + x^{2}} dx = \tan^{- 1} (x) + C$
$f ‘ (x) = \frac{1}{\| x \| \sqrt{x^{2} - 1}}$	$\int_{a}^{b} \frac{1}{\| x \| \sqrt{x^{2} - 1}} dx = {\sec^{- 1} (x) \|}_{x = a}^{x = b}$	$\int \frac{1}{\| x \| \sqrt{x^{2} - 1}} dx = \sec^{- 1} (x) + C$
$f ‘ (x) = \frac{- 1}{\| x \| \sqrt{x^{2} - 1}}$	$\int_{a}^{b} \frac{- 1}{\| x \| \sqrt{x^{2} - 1}} dx = {\csc^{- 1} (x)}_{x = a}^{x = b}$	$\int \frac{- 1}{\| x \| \sqrt{x^{2} - 1}} dx = \csc^{- 1} (x) + C$
$f ‘ (x) = \frac{- 1}{1 + x^{2}}$	$\int_{a}^{b} \frac{- 1}{1 + x^{2}} dx = {\cot^{- 1} (x) \|}_{x = a}^{x = b}$	$\int \frac{- 1}{1 + x^{2}} dx = \cot^{- 1} (x) + C$

Note : As soon as whatever is inside the function, inside the parenthesis, is more than just your basic x , the problem’s primary method becomes a u-substitution .

Conveniently math teachers have not come to an agreement on how to refer to inverse trig functions. Depending on where you are taking your AP Calculus class you could see either version, (i.e., $\sin^{- 1} (x)$ or $arc sin (x)$ ) used to describe an inverse trig function. They both mean the same thing, and they both have the same derivative.