Here we look at the integration of basic trig functions. These results can be extended by application to integration by substitution and integration by parts.

## Integration of Trigonometric Functions

As an immediate consequence of the results on differentiation, we have $$ \int \sin(x)dx = -\cos(x) + c $$ $$ \int \cos(x)dx = \sin(x) + c $$ Application of a substitution yields $$ \int \sin(kx)dx = -\frac{1}{k}\cos(x) + c $$ $$ \int \cos(kx)dx = \frac{1}{k}\sin(kx) + c $$

## Integration of Trigonometric Functions 1

Evaluate the following integral $$ \int \sin(2t)dt $$

solution - press button to display

$$ \int\sin(2t)dt = -\frac{1}{2}\cos(2t) + c $$

## Integration of Trigonometric Functions 2

Evaluate the following integral $$\int \sin(3t)\cos(3t)dt $$

solution - press button to display

The key trick in this integration is the application of the double angle formula. $$\begin{align} \int \sin(3t)\cos(3t)dt&= \int \frac{1}{2}\sin(6t)dt \\ &= -\frac{1}{12}\cos(6t)+ c \end{align} $$

## Integration of Trigonometric Functions 3

Evaluate the integral $$ \int \sin t \cos^3t dt $$

solution - press button to display

Let $u = \cos t$ then $du = -\sin t dt$. Substitution of these terms yields $$ \int \sin (t) \cos^3(t) dt = \int -u^3du = -\frac{1}{4}u^{4} + c $$