# Integrals of Trigonometric Functions

## Key Questions

• Recall:

$\int \frac{g ' \left(x\right)}{g \left(x\right)} \mathrm{dx} = \ln | g \left(x\right) | + C$

(You can verify this by substitution $u = g \left(x\right)$.)

Now, let us look at the posted antiderivative.

By the trig identity $\tan x = \frac{\sin x}{\cos x}$,

$\int \tan x \mathrm{dx} = \int \frac{\sin x}{\cos x} \mathrm{dx}$

by rewriting it a bit further to fit the form above,

$= - \int \frac{- \sin x}{\cos x} \mathrm{dx}$

by the formula above,

$= - \ln | \cos x | + C$

or by $r \ln x = \ln {x}^{r}$,

$= \ln | \cos x {|}^{- 1} + C = \ln | \sec x | + C$

I hope that this was helpful.

• Since

$\left(\tan x\right) ' = {\sec}^{2} x$,

we have

$\int {\sec}^{2} x \mathrm{dx} = \tan x + C$.

I hope that this was helpful.

• Since

$\left(\ln | \sec x + \tan x |\right) ' = \frac{\sec x \tan x + {\sec}^{2} x}{\sec x + \tan x} = \sec x$,

we have

$\int \sec x \mathrm{dx} = \ln | \sec x + \tan x | + C$

Since

$\left(- \ln | \csc x + \cot x |\right) ' = - \frac{- \csc x \cot x - {\csc}^{2} x}{\csc x + \cot x} = \csc x$,

we have

$\int \csc x \mathrm{dx} = - \ln | \csc x + \cot x | + C$

$\int \cot x \mathrm{dx} = \int \frac{\cos x}{\sin x} \mathrm{dx} = \ln | \sin x | + C$

I hope that this was helpful.

• Since

$\left(\sin x\right) ' = \cos x$,

we have

$\int \cos x \mathrm{dx} = \sin x + C$.

Since

$\left(- \cos x\right) ' = \sin x$,

we have

$\int \sin x \mathrm{dx} = - \cos x + C$.

I hope that this was helpful.