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.
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.
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.
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!
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
For this example, it was necessary that \sec had an even power so we could factor out a \sec^2 and convert the rest of our \sec to \tan functions.
Problems
I am running out of time to prepare for the competition, so the problems listed may be simpler then what to expect for the integration bee.
Integrals involving powers of trigonometric functions
Evaluate the following integral
\int \sec^9 x \tan^5 x \, dx
Evaluate the following integral
\int \sec^3 x \tan^5 x \, dx
Evaluate the following integral
\int \sec^4 x \tan^6 x \, dx
Evaluate the following integral
\int \sin^8 x \cos^5 x \, dx