Definite integrals of trigonometric functions

The antiderivative of $\sin x$ is $-\cos x$ . We use it to find the area of one arch of the sine curve.

\int_0^{\pi} \sin x \, dx = \left[ -\cos x \right]_0^{\pi} = -(-1) - (-1) = 2

The area is $2$ . The arch has height $1$ and width $\pi \approx 3.14$ , so against the rectangle of the same width and height, of area $\pi$ , the ratio is $\dfrac{2}{\pi} \approx 0.64$ . An arch of the sine fills about six tenths of the rectangle.

Mind the sign

Function	Antiderivative
$\sin x$	$-\cos x$
$\cos x$	$\sin x$

Since $(-\cos x)' = \sin x$ , the antiderivative is $-\cos x$ and not $\cos x$ . It is the flip side of the minus sign that appears when $\cos$ is differentiated into $-\sin$ .

Over a full period it is zero

\int_0^{2\pi} \sin x \, dx = \left[ -\cos x \right]_0^{2\pi} = -1 - (-1) = 0

It comes out $0$ because the trough from $\pi$ to $2\pi$ lies below the $x$ -axis and is counted negative.

Interval	Definite integral	Area
$0$ to $\pi$	$2$	$2$
$\pi$ to $2\pi$	$-2$	$2$
$0$ to $2\pi$	$0$	$4$

To get the area $4$ , integrate $|\sin x|$ , or double the $2$ obtained from a single arch.

Reading it on the graph of the antiderivative

The slope of $y = -\cos x$ is $0$ at $x = 0$ , $1$ at $x = \dfrac{\pi}{2}$ and $0$ at $x = \pi$ , which are exactly the values of $\sin x$ . Where $\sin$ is positive $-\cos$ increases, and where $\sin$ is negative it decreases.

The rise from $-\cos 0 = -1$ to $-\cos \pi = 1$ , an increase of $2$ , is the area itself. That a definite integral is a difference of antiderivatives can be seen between the two graphs.

The wave on the graph is $y = \sin x$ , the wave lagging it by $\dfrac{\pi}{2}$ is the antiderivative $y = -\cos x$ , and the large dots are the two ends of the integral.