Trigonometric Identities and Powers

We now examine various trigonometric identities that will help us in our quest to win the integration bee, and examine how to deal with powers of trigonometric functions.

Identities

We start by looking at various forms of 1.

Theorem (Pythagorean Identities)

The following are true

\begin{align*} \sin^2 x + \cos^2 x &= 1\\ \sec^2 x - \tan^2 x &= 1\\ \csc^2 x - \cot^2 x &= 1 \end{align*}

Of course, these identities can be rearranged when needed, so we’ll often use identities like \tan^2 x - 1 = \sec^2 x. Next we examine double angle identities.

Theorem (Double angle identities)

The following double angle identities hold

\begin{align*} \sin(2x) &= 2\sin x \cos x\\ \cos(2x) &= 2\cos^2x - 1\\ \cos(2x) &= 1 - 2\sin^2 x \end{align*}

and can be rewritten

\begin{align*} \sin x \cos x &= \frac{1}{2}\sin(2x)\\ \cos^2x &= \frac{1}{2}(1 + \cos(2x))\\ \sin^2x &= \frac{1}{2}(1 - \cos(2x)) \end{align*}

There also exist half-angle identities and formulas for products of angles. [TO-DO]

Integrals of powers of trigonometric functions

The motivating example for this section is the integral

\int \sin^5 x \, dx

which, although it can be solved by integration by parts, is much simpler if we use the identity \sin^2 = 1 - \cos^2.

Example (Powers of sin)

Given the following integral \int \sin^5 x \, dx

We take out a \sin x and use our identity to obtain

\int \left(1-\cos^2 x\right)^2 \sin x \, dx

Using u = \cos x we obtain

-\int \left(1-u^2 \right)^2 \, du

whereby inverse power rule gets us

= -\left(u - \frac{2}{3}u^3 + \frac{1}{5}u^5\right) + C

and reversing the u-sub gives us

= -\cos x + \frac{2}{3}\cos^3 x - \frac{1}{5}\cos^5 x + C

This is the general method when we have an odd exponent on either \sin or \cos. We factor out one exponent, use identities to transform all but our factored exponent into some other function, and then preform a u-sub.

Lets see how this looks like with \tan and \sec functions!

Example (Powers of sec)

Given the following integral \int \sec^6 x \, dx

We take out a \sec^2 x and use our identity to obtain

\int \left(\tan^2 x + 1\right)^2 \sec^2 x \, dx

Using u = \tan x we obtain

\int \left(u^2 + 1 \right)^2 \, du

whereby inverse power rule gets us

= \left(\frac{1}{5}u^5\right + \frac{2}{3}u^3) + C

and reversing the u-sub gives us

= \frac{1}{5}\tan^5 x + \frac{2}{3}\tan^3 x + C