Integrating Trig Functions

Exam code: 4037

Written by: Ashika|Reviewed by: Caroline Carroll|Updated 2 July 2026

Integrating trig functions

$\int \sin x d x = - \cos x + c$

$\int \cos x d x = \sin x + c$

where $c$ is the constant of integration

$\int \sec^{2} x d x = \tan x + c$

$\int \sin (a x + b) d x = - \frac{1}{a} \cos (a x + b) + c$

$\int \cos (a x + b) d x = \frac{1}{a} \sin (a x + b) + c$

$\int \sec^{2} (a x + b) d x = \frac{1}{a} \tan (a x + b) + c$

For calculus with trigonometric functions angles must be measured in radians

Worked example

a) Find, in the form $f (x) + c$ , an expression for each integral

i. $\int \cos x d x$

ii. $\int \sec^{2} (3 x - \frac{π}{3}) d x$

i. This is a result you should be able to recall from memory.

$\int \cos x d x = \sin x + c$

ii. Use the standard result: $\int \sec^{2} (a x + b) d x = \frac{1}{a} \tan (a x + b) + c .$

$\int \sec^{2} (3 x - \frac{π}{3}) d x = \frac{1}{3} \tan (3 x - \frac{π}{3}) + c$

b) A curve has the following equation:

$y = \int 2 \sin (2 x + \frac{π}{6}) d x$

The curve passes through the point with coordinates $(\frac{π}{3}, \sqrt{3}) .$

Find an expression for $y .$

Factor out the 2 in front of the sin first, and then use the standard result: $\int \sin (a x + b) d x = - \frac{1}{a} \cos (a x + b) + c .$

$y = 2 \int \sin (2 x + \frac{π}{6}) d x y = 2 (- \frac{1}{2} \cos (2 x + \frac{π}{6}) + c)$

When multiplying the arbitrary constant $c$ , by 2, it is still just an arbitrary constant and so it is still just written as $c .$

$y = - \cos (2 x + \frac{π}{6}) + c$

Substitute in the given coordinate, and evaluate, to find $c .$

At $x = \frac{π}{3}, y = \sqrt{3}$

$\sqrt{3} = - \cos (\frac{2 π}{3} + \frac{π}{6}) + c \sqrt{3} = - \cos (\frac{5 π}{6}) + c \sqrt{3} = \frac{\sqrt{3}}{2} + c c = \frac{\sqrt{3}}{2}$

Rewrite the full expression for $y .$

$y = - \cos (2 x + \frac{π}{6}) + \frac{\sqrt{3}}{2}$