Integrals of common functions

This section introduces the rules for trigonometric identities.

Trig identities

There are six commonly used trigonometric functions, all of which can be expressed using only two, \sin and \cos. They are defined as follows.

Definition

Let \sin(x) and \cos(x) be our starting functions. Then we define the following functions

The Tangent function defined

\tan(x) = \frac{\sin(x)}{\cos(x)}

The Secant function defined

\sec(x) = \frac{1}{\cos(x)}

The Cosecant function defined

\csc(x) = \frac{1}{\sin(x)}

and the Cotangent function defined

\cot(x) = \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)}

And of course each trigonmetric identity has a derivative, and a corresponding integral

Definition

The derivatives of each trigonometric identity are as follows.

\begin{array}{ll} \sin^\prime(x) = \cos(x) & \cos^\prime(x) = -\sin(x)\\ \tan^\prime(x) = \sec^2(x) \hspace{0.5in} & \cot^\prime(x) = -\csc^2(x)\\ \sec^\prime(x) = \sec(x)\tan(x) \hspace{0.5in} & \csc^\prime(x) = -\csc(x)\cot(x) \end{array}

Which give us the following integrals

\begin{array}{ll} \displaystyle\int{{\cos x\,dx}} = \sin x + C \hspace{0.5in} & \displaystyle\int{{\sin x\,dx}} = - \cos x + C\\ \displaystyle\int{{{{\sec }^2}x\,dx}} = \tan x + C \hspace{0.5in} & \displaystyle\int{{{{\csc }^2}x\,dx}} = - \cot x + C\\ \displaystyle\int{{\sec x\tan x\,dx}} = \sec x + C \hspace{0.5in} & \displaystyle\int{{\csc x\cot x\,dx}} = - \csc x + C \end{array}

It is helpful (for me at least) to memorize the left column of the table, and then derive the right side by replacing each function with its ‘complement’, and then negating the result.

Inverse Trig identities

Each trig function has an inverse function, which may be helpful in our integrals. We traditionally only look at the inverses of \sin,\tan and \sec.

Theorem

The following integrals evaluate to inverse trig functions,

\int \frac{1}{x^2 + 1} \, dx = \arctan(x) + C

\int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin(x) + C

\int \frac{1}{x\sqrt{x^2 - 1}}\,dx = \text{arcsec}(x) + C

It is worth noting that the derivative of \arccos is very similar to \arcsin \frac{d}{dx}\arccos(x) = -\frac{1}{\sqrt{1-x^2}}.

So our second integral can also be solved

\int \frac{1}{\sqrt{1-x^2}}\,dx = -\arccos(x) + C.

This is because \arcsin and -\arccos differ only by a constant (\frac{\pi}{2}).

Exponential and logarithmic functions

By once again inverting common derivative rules, we get the following identities.

Theorem

\int e^x \, dx = e^x + C \hspace{0.5in} \int a^x \,dx = \frac{a^x}{\ln{a}} + C