Six Trig Functions

Unit 6: Antiderivatives Six Trig Functions

Base Case: Six Trig Functions

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

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 .