Integrating with Trig Identities

In short: Some trig integrals need an identity first. Use $sin^{2} x = \frac{1}{2} (1 - cos 2 x)$ and $cos^{2} x = \frac{1}{2} (1 + cos 2 x)$ to integrate squares, and $tan^{2} x = sec^{2} x - 1$ for $tan^{2} x$ .

Starting point: what you already know

Double angle formulas (Compound Angles and Double Angles): $cos 2 x = 1 - 2 sin^{2} x = 2 cos^{2} x - 1$ and $sin 2 x = 2 sin x cos x$ .
Pythagorean identities (Reciprocal Trigonometric Functions): $sin^{2} x + cos^{2} x = 1$ and $1 + tan^{2} x = sec^{2} x$ .
Integrals of trig functions from Mathematics Advanced: $\int cos a x d x = \frac{1}{a} sin a x + C$ , $\int sin a x d x = - \frac{1}{a} cos a x + C$ , $\int sec^{2} a x d x = \frac{1}{a} tan a x + C$ .
Radians: calculus with trig always uses radians.

Rewrite the squares

Rearrange the double angle formulas for $cos 2 x$ :

Now each one is easy to integrate:

$\int sin^{2} x d x = \frac{x}{2} - \frac{sin 2 x}{4} + C \int cos^{2} x d x = \frac{x}{2} + \frac{sin 2 x}{4} + C$

y = sin² x swings between 0 and 1. The humps above the dashed line exactly fill the dips below it, so the average is one half.

Other products and powers

Integrating with trig identities

Squares

sin² x = ½(1 - cos 2x)
cos² x = ½(1 + cos 2x)

Products

sin x cos x = ½ sin 2x

tan² x

sec² x - 1
Gives tan x - x

Inside angle

sin² 3x uses cos 6x

Uses

Average value
Volumes of solids

Quick summary

$sin^{2} x = \frac{1}{2} (1 - cos 2 x)$ and $cos^{2} x = \frac{1}{2} (1 + cos 2 x)$ .
$\int sin^{2} x d x = \frac{x}{2} - \frac{sin 2 x}{4} + C$ and $\int cos^{2} x d x = \frac{x}{2} + \frac{sin 2 x}{4} + C$ .
$sin x cos x = \frac{1}{2} sin 2 x$ and $tan^{2} x = sec^{2} x - 1$ .
For $sin^{2} k x$ , the new angle is $2 k x$ .
Odd powers: keep one factor and substitute.

Quiz

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Try 3 questions now

Which identity rewrites $cos^{2} x$ ?
1. A $\frac{1}{2} (1 - cos 2 x)$
2. B $\frac{1}{2} (1 + cos 2 x)$
3. C $1 - sin 2 x$
4. D $2 cos 2 x - 1$
Show answer
B: $\frac{1}{2} (1 + cos 2 x)$
From $cos 2 x = 2 cos^{2} x - 1$ .
Which identity rewrites $sin^{2} x$ ?
1. A $\frac{1}{2} (1 - cos 2 x)$
2. B $\frac{1}{2} (1 + cos 2 x)$
3. C $\frac{1}{2} sin 2 x$
4. D $1 + cos^{2} x$
Show answer
A: $\frac{1}{2} (1 - cos 2 x)$
From $cos 2 x = 1 - 2 sin^{2} x$ .
Find $\int cos 2 x d x$ .
1. A $2 sin 2 x + C$
2. B $\frac{1}{2} sin 2 x + C$
3. C $- \frac{1}{2} sin 2 x + C$
4. D $sin 2 x + C$
Show answer
B: $\frac{1}{2} sin 2 x + C$
Divide by the 2 inside.