8.3a. Use of trigonometric identities

The Pythagorean identities

We recall the Pythagorean identities
$\sin^2 A + \cos^2 A = 1 \\ \tan^2 A + 1 = \mathrm{sec}^2 A \\ 1 + \mathrm{cot}^2 A = \mathrm{cosec}^2 A$

Since we know $\displaystyle \int \mathrm{sec}^2 x \; \mathrm{d}x = \tan x +C$, we can use the second identity to solve $\displaystyle \int \tan^2 x \; \mathrm{d}x = \int \mathrm{sec}^2 x - 1 \; \mathrm{d}x = \tan x - x + C$

The double angle formulas

Before we proceed, let us first take a look at a common mistake. $\displaystyle \int \sin^2 x \; \mathrm{d}x$ IS NOT $\displaystyle \frac{sin^3 x}{3}$. As seen in the discussion for example 8.0 and example 8.1 , we need the presence of an "$f'(x)$" to use the $\int x \; \mathrm{d}x = \frac{x^2}{2}$ formula.

The Pythagorean identities aren't especially useful for $\displaystyle \int \sin^2 x \; \mathrm{d}x$ since we do not know how to integrate $\cos^2 x$ either. Instead, we will use the double angle formula:
$\cos 2A \equiv \cos^2 A - \sin^2 A \equiv 2 \cos^2 A - 1 \equiv 1 - 2 \sin^2 A$

We rearrange the double angle formula to make $\sin^2 x$ the subject to get $\displaystyle \int \sin^2 x \; \mathrm{d}x = \int \frac{1-\cos 2x}{2} \; \mathrm{d}x = \frac{x}{2} - \frac{\sin 2x}{4} + C$.

Trigonometric integrals given in MF26

We should also be familiar with the integrals provided in the formula list so that we can use them when necessary. Similar to the domain considerations for our logarithm integrals in examples 8.0 , 8.1 and 8.2 , the formulas here have an extra modulus compared to the one given on the formula sheet. Note that arbitrary constants are omitted for the following:
$\displaystyle \int \tan x \; \mathrm{d}x = \ln | \mathrm{sec} \, x | \\ \displaystyle \int \mathrm{cot} \, x \; \mathrm{d}x = \ln | \sin x | \\ \displaystyle \int \mathrm{cosec} \, x \; \mathrm{d}x = - \ln | \mathrm{cosec} \, x + \mathrm{cot} \, x | \\ \displaystyle \int \mathrm{sec} \, x \; \mathrm{d}x = \ln | \mathrm{sec} \, x + \tan x |$

More practice

  1. Use the double angle formula to find $\displaystyle \int \cos^2 3x \; \mathrm{d}x$.
  2. Use the result $\int \sin^2 x \; \mathrm{d}x = \frac{x}{2}-\frac{\sin 2x}{4} +C$ we have found in the earlier discussion to find $\displaystyle \int \cos^2 x \; \mathrm{d}x$.
  3. Use the double angle formula to find $\displaystyle \int \frac{1}{1+\cos 2x} \; \mathrm{d}x$.
  4. Use the double angle formula to find $\displaystyle \int \sqrt{1-\cos 2x} \; \mathrm{d}x$ for $0 \leq x \leq \pi$.
  5. Use the sine double angle formula to find $\displaystyle \int \sin x \cos x \; \mathrm{d}x$.
Solutions to "more practice"

1. $\displaystyle \int \cos^2 3x \; \mathrm{d}x = \int \frac{\cos 6x + 1}{2} \; \mathrm{d}x = \frac{\sin 6x}{12} + \frac{x}{2} + C.$

2. $\displaystyle \int \cos^2 x \; \mathrm{d}x = \int 1-\sin^2 x \; \mathrm{d}x = x - \left (\frac{x}{2} - \frac{\sin 2x}{4} + C \right ) = \frac{x}{2} + \frac{\sin 2x}{4} + C'$

Questions 3 and 4 can be tricky because the presence of the $1$ alongside the cosine function stops us from manipulating the expression. We thus apply the appropriate double angle formula "forward" to eliminate $1$.
3. $\displaystyle \int \frac{1}{1 + \cos 2x} \; \mathrm{d}x = \int \frac{1}{1 + (2 \cos^2 x - 1)} \; \mathrm{d}x = \int \frac{1}{2} \mathrm{sec}^2 x \; \mathrm{d}x = \frac{1}{2} \tan x + C$.
4. $\displaystyle \int \sqrt{1-\cos 2x} \; \mathrm{d}x = \int \sqrt{1 - (1 - 2 \sin^2 x)} \; \mathrm{d}x = \int \sqrt{2} \sin x \; \mathrm{d}{x} = -\sqrt{2} \cos x + C$.

Question 5 uses the sine double angle formula $\sin 2A \equiv 2 \sin A \cos A$. We have
5. $\displaystyle \int \sin x \cos x \; \mathrm{d}x = \int \frac{\sin 2x}{2} \; \mathrm{d} x = \frac{-\cos 2x}{4} + C$.