Integrating trigonometric functions

Standard integrals of sin, cos and sec squared, definite integrals, finding a primitive from an initial condition, the integral of tan x, and the reverse chain rule with trigonometric functions.

Standard forms

Trigonometric integrals

$$\int \cos(ax + b)\,dx = \frac1a\sin(ax + b) + C \qquad \int \sin(ax + b)\,dx = -\frac1a\cos(ax + b) + C \qquad \int \sec^2(ax + b)\,dx = \frac1a\tan(ax + b) + C$$

Example 1●

Evaluate (a) $\displaystyle\int_0^{\pi} \sin x\,dx$ and (b) $\displaystyle\int_0^{\frac{\pi}{6}} \cos 2x\,dx$.

$\Big[-\cos x\Big]_0^{\pi} = 1 - (-1) = 2$

$\Big[\tfrac12\sin 2x\Big]_0^{\frac{\pi}{6}} = \tfrac12\sin\dfrac{\pi}{3} = \dfrac{\sqrt3}{4}$

Example 2: an initial condition●●

$f'(x) = \sin 2x$ and $f(0) = 1$. Find $f(x)$.

$f(x) = -\tfrac12\cos 2x + C$, and $f(0) = -\tfrac12 + C = 1$, so $C = \tfrac32$.

$f(x) = \tfrac32 - \tfrac12\cos 2x$

The reverse chain rule

Example 3●●

Find (a) $\displaystyle\int \tan x\,dx$, (b) $\displaystyle\int 2x\cos x^2\,dx$ and (c) $\displaystyle\int \sin x\cos^2 x\,dx$.

$\tan x = \dfrac{\sin x}{\cos x} = -\dfrac{-\sin x}{\cos x}$, which is $-\dfrac{f'(x)}{f(x)}$. So the integral is $-\ln|\cos x| + C$.

$2x$ is the derivative of $x^2$: $\sin x^2 + C$.

$-\sin x$ is the derivative of $\cos x$: $-\displaystyle\int (-\sin x)(\cos x)^2\,dx = -\dfrac{\cos^3 x}{3} + C$.

Flashcards

$displaystyleint sin ax,dx$

$-\dfrac1a\cos ax + C$

$displaystyleint cos ax,dx$

$\dfrac1a\sin ax + C$

$displaystyleint an x,dx$

$-\ln|\cos x| + C$

Practice

$\displaystyle\int \cos 3x\,dx = $
$\displaystyle\int \sin 2x\,dx = $
Evaluate $\displaystyle\int_0^{\pi} \sin x\,dx$.
Evaluate $\displaystyle\int_0^{\frac{\pi}{4}} \sec^2 x\,dx$.
Evaluate $\displaystyle\int_0^{\frac{\pi}{6}} \cos 2x\,dx$. Type $\sqrt3$ as sqrt3.
$f'(x) = \sin 2x$ and $f(0) = 1$. Find $f\left(\tfrac{\pi}{2}\right)$.
Evaluate $\displaystyle\int_0^{\frac{\pi}{3}} \tan x\,dx$. Type $\ln 2$ as ln2.
$\displaystyle\int 2x\cos x^2\,dx = $
$\displaystyle\int \sin x\cos^2 x\,dx = $