Indefinite Integrals of Trigonometric Functions

# Indefinite Integrals of Trigonometric Functions

The indefinite integrals for $\sin x$ and $\cos x$ should be rather easy to remember as we are just going in the opposite cyclic pattern for differentiation. Recall that the cycle of $\sin x \to \cos x \to -\sin x \to -\cos x \to \sin x \to ...$ is a good memory device for remembering the derivatives of a function. For example, the derivative of $\sin x$ is $\cos x$ (by following the arrows). For integration, the same mechanic works just in the opposite direction, so the antiderivative of $\cos x$ is $\sin x$. This information is summarized in the following property:

 Theorem 1: The following functions have the following indefinite integrals: a) If $f(x) = \sin x$, then $\int \sin x \: dx = -\cos x + C$. b) If $f(x) = \cos x$, then $\int \cos x \: dx = \sin x + C$. c) If $f(x) = \tan x$, then $\int \tan x \: dx = \ln \mid \sec x \mid + C$.

The indefinite integral for $\tan x$ is a little more complicated, and we cannot verify this integral yet. Let's look at some examples.

## Example 1

Evaluate the following integral: $\int 5 \cos x \: dx$.

From the Properties of Indefinite Integrals, and from the rules above, we can simplify this:

(1)
\begin{align} \int 5 \cos x \: dx \\ = 5 \int \cos x \: dx \\ = 5 \sin x + C \end{align}

## Example 2

Evaluate the following integral: $\int 4t^3 + \sin t \: dt$.

Once again, we will apply what we know:

(2)
\begin{align} \int 4t^3 + \sin t \: dt \\ = \int 4t^3 \: dt + \int \sin t \: dt \\ = \frac{4t^{4}}{4} -\cos t + C \\ = t^4 - \cos t + C \end{align}

## Example 3

Evaluate the following integral: $\int \frac{\sin x}{\cos x} + \tan x - x^\pi \: dx$.

Recall that $\frac{\sin x}{\cos x} = \tan x$, so it follows that:

(3)
\begin{align} \int \frac{\sin x}{\cos x} + \tan x - x^\pi \: dx = \int \tan x + \tan x - x^\pi \: dx \end{align}

Now we can integrate:

(4)
\begin{align} \int \tan x + \tan x - x^\pi \: dx \\ = \int 2 \tan x - x^\pi \: dx \\ = \int 2 \tan x \: dx - \int x^\pi \: dx \\ = 2 \int \tan x \: dx - \int x^\pi \: dx \\ = 2 \ln |\sec x | - \frac{x^{\pi + 1}}{\pi + 1} + C \\ \end{align}