Integrating sin² and cos²

Why $\sin^{2}$ needs a rewrite

The four functions on the sine and cosine ring differentiate to sin x, cos x, −sin x and −cos x. None of them differentiates to $\sin^{2}x$, so there is no integral to read off the ring.

Treating sin x as if it were x does not work either. The guess $\frac{\sin^{3}x}{3}$ differentiates, by the chain rule, to $\frac{{3 \sin^{2}x \times \cos x}}{3} = \sin^{2}x \cos x$. That has an extra factor of cos x, so $\frac{\sin^{3}x}{3}$ is not the integral of $\sin^{2}x$.

The double angle formula removes the square

One form of the double angle formula for cosine is $\cos 2x = 1 - 2 \sin^{2}x$. Rearrange it for the square: $2 \sin^{2}x = 1 - \cos 2x$, so $\sin^{2}x = \frac{{1 - \cos 2x}}{2}$.

The right-hand side has no square in it. It is a constant, ½, minus a plain cosine, ½ cos 2x, and both of those integrate directly.

Check at $x = \frac{\pi}{6}$: $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$, so $\sin^{2}x = \frac{1}{4}$. On the right, $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$, so $\frac{{1 - \frac{1}{2}}}{2} = \frac{1}{4}$ as well.

The curve $y = \sin^{2}x$ and the line y = ½. The curve swings between 0 and 1, an amount ½ either side of the line, and repeats every $\pi$, which is the period of cos 2x. That is what $\sin^{2}x = \frac{1}{2} - \frac{1}{2} \cos 2x$ says: a constant ½, and a wave of size ½ about it.

Integrating the pieces

Now $\int \sin^{2}x \,dx = \int \left(\frac{1}{2} - \frac{1}{2} \cos 2x\right) \,dx$. The ½ integrates to $\frac{x}{2}$. The cos 2x has a 2 inside, so it integrates to $\frac{{\sin 2x}}{2}$, and ½ of that is $\frac{{\sin 2x}}{4}$. Together, $\int \sin^{2}x \,dx = \frac{x}{2} - \frac{{\sin 2x}}{4} + C$.

Check by differentiating. $\frac{x}{2}$ gives ½, and $\frac{{\sin 2x}}{4}$ gives $\frac{{2 \cos 2x}}{4} = \frac{1}{2} \cos 2x$. So the derivative is ½ − ½ cos 2x, which is $\sin^{2}x$.

$\cos^{2}$ takes a plus

Another form of the same formula is $\cos 2x = 2 \cos^{2}x - 1$. Rearranged, $\cos^{2}x = \frac{{1 + \cos 2x}}{2}$: the same move, with a plus where $\sin^{2}$ had a minus. Check at x = 0: $\cos^{2}0 = 1$, and $\frac{{1 + 1}}{2} = 1$.

So $\int \cos^{2}x \,dx = \frac{x}{2} + \frac{{\sin 2x}}{4} + C$. Adding the two results gives $\int \left(\sin^{2}x + \cos^{2}x\right) \,dx = x + C$, which is right, since $\sin^{2}x + \cos^{2}x = 1$.

The area under $y = \cos^{2}x$ from 0 to X, measured against the line y = ½. The term $\frac{X}{2}$ is the rectangle under the line; the term $\frac{{\sin 2X}}{4}$ is what the curve adds above the line less what it leaves out below. At X = 2.2 that is 1.1 − 0.238 = 0.862. Drag X to $\frac{\pi}{2}$, $\pi$, $\frac{3\pi}{2}$ or $2\pi$, where sin 2X = 0: the parts above and below the line cancel, and the area is exactly $\frac{X}{2}$.

Definite integrals

Over half a turn the wave term vanishes. The integral of $\sin^{2}x$ from 0 to $\pi$ is $\left[\frac{x}{2} - \frac{{\sin 2x}}{4}\right]_{0}^{\pi} = \frac{\pi}{2} - 0$, since $\sin 2\pi = 0$ and sin 0 = 0. The integral of $\cos^{2}x$ over the same interval is also $\frac{\pi}{2}$, and the two add to $\pi$, the integral of 1.

Over a part of a wave it does not. The integral of $\sin^{2}x$ from 0 to $\frac{\pi}{4}$ is $\frac{\pi}{8} - \frac{{\sin\left(\frac{\pi}{2}\right)}}{4} = \frac{\pi}{8} - \frac{1}{4} \approx 0.143$. That is less than half of the width $\frac{\pi}{4} \approx 0.785$, because $\sin^{2}x$ stays below ½ until $x = \frac{\pi}{4}$.

Other angles and products

With 3x in place of x, the double angle is 6x: $\sin^{2}\left(3x\right) = \frac{{1 - \cos 6x}}{2}$, so $\int \sin^{2}\left(3x\right) \,dx = \frac{x}{2} - \frac{{\sin 6x}}{12} + C$. Check: the derivative is $\frac{1}{2} - \frac{{6 \cos 6x}}{12} = \frac{1}{2} - \frac{1}{2} \cos 6x$.

A product of sine and cosine uses the sine formula instead. sin 2x = 2 sin x cos x, so $\sin x \cos x = \frac{{\sin 2x}}{2}$, which integrates to $- \frac{{\cos 2x}}{4} + C$. Check: $- \frac{{\cos 2x}}{4}$ differentiates to $\frac{{2 \sin 2x}}{4} = \frac{{\sin 2x}}{2}$.

The usual mistakes

Integrating the square as a power. $\frac{\sin^{3}x}{3}$ differentiates to $\sin^{2}x \cos x$, not $\sin^{2}x$.

The plus and minus swapped. $\sin^{2}x = \frac{{1 - \cos 2x}}{2}$ has the minus. At x = 0, $\sin^{2}x$ is 0, and only the minus version gives $\frac{{1 - 1}}{2} = 0$.

Halving one piece and not the other. Both the 1 and the cos 2x are divided by 2, so the answer is $\frac{x}{2} - \frac{{\sin 2x}}{4} + C$, not $x - \frac{{\sin 2x}}{4} + C$.

Forgetting the 2 inside. cos 2x integrates to $\frac{{\sin 2x}}{2}$, not sin 2x; with the ½ in front, that makes $\frac{{\sin 2x}}{4}$.

Using $\sin^{2}x + \cos^{2}x = 1$. It only trades $\sin^{2}x$ for $1 - \cos^{2}x$, which still has a square.