Trigonometry is a huge part of the integration bee, and will repeatedly appear in a large portion of integrals. As such it is crucial to have a complete understanding of these functions and how they interact.
Trigonometric Functions
There are six commonly used trigonometric functions, all of which can be expressed using only two, \sin and \cos. They are defined as follows.
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 function has a derivative, and a corresponding integral
The derivatives of each trigonometric function 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 Trigonometric functions
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.
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}).
Hyperbolic Trigonometric functions
For each trigonometric function, we can take it’s hyperbolic counterpart.
The derivatives of each hyperbolic trigonometric function are as follows.
\begin{array}{ll}
\sinh^\prime(x) = \cosh(x) & \cosh^\prime(x) = \sinh(x)\\
\tanh^\prime(x) = \text{sech}^2(x) \hspace{0.5in} & \text{coth}^\prime(x) = -\text{csch}^2(x)\\
\text{sech}^\prime(x) = -\text{sech}(x)\tanh(x) \hspace{0.5in} & \text{csch}^\prime(x) = -\text{csch}(x)\coth(x)
\end{array}
Which give us the following integrals
\begin{array}{ll}
\displaystyle\int{{\cosh x\,dx}} = \sinh x + C \hspace{0.5in} & \displaystyle\int{{\sinh x\,dx}} = \cosh x + C\\
\displaystyle\int{{{{\text{sech} }^2}x\,dx}} = \tanh x + C \hspace{0.5in} & \displaystyle\int{{{{\text{csch} }^2}x\,dx}} = - \cot x + C\\
\displaystyle\int{{\text{sech} x\tanh x\,dx}} = -\text{sech} x + C \hspace{0.5in} & \displaystyle\int{{\text{csch} x\,\text{coth}\, x\,dx}} = - \text{csch} x + C
\end{array}
My strategy for these is just to hope they don’t show up.
Inverse Hyperbolic Trigonometric functions
But wait, theres more!
The following integrals evaluate to inverse trig functions,
\int \frac{1}{1 - x^2} \, dx = \text{arctanh}(x) + C
\int \frac{1}{\sqrt{x^2 + 1}}\,dx = \text{arcsinh}(x) + C
\int \frac{-1}{x\sqrt{1 - x^2}}\,dx = \text{arcsech}(x) + C