Integrating sin² and cos²

Why ${\sin }^2$sin² 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 ${\sin }^{2}x$sin²x, so there is no integral to read off the ring.

Treating sin x as if it were x does not work either. The guess $\frac{{\sin }^{3}x}{3}$sin³x/3 differentiates, by the chain rule, to $\frac{3{\sin }^{2}x\times \cos x}{3}={\sin }^{2}x\cos x$(3 sin²x × cos x)/3 = sin²x cos x. That has an extra factor of cos x, so $\frac{{\sin }^{3}x}{3}$sin³x/3 is not the integral of ${\sin }^{2}x$sin²x.

The double angle formula removes the square

One form of the double angle formula for cosine is $\cos 2x=1-2{\sin }^{2}x$cos 2x = 1 − 2 sin²x. Rearrange it for the square: $2{\sin }^{2}x=1-\cos 2x$2 sin²x = 1 − cos 2x, so ${\sin }^{2}x=\frac{1-\cos 2x}{2}$sin²x = (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 $x=\frac{\pi }{6}$x = π/6: $\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}$sin(π/6) = ½, so ${\sin }^{2}x=\frac{1}{4}$sin²x = 1/4. On the right, $\cos \left(\frac{\pi }{3}\right)=\frac{1}{2}$cos(π/3) = ½, so $\frac{1-\frac{1}{2}}{2}=\frac{1}{4}$(1 − ½)/2 = 1/4 as well.

The curve $y={\sin }^{2}x$y = sin²x and the line y = ½. The curve swings between 0 and 1, an amount ½ either side of the line, and repeats every $\pi$π, which is the period of cos 2x. That is what ${\sin }^{2}x=\frac{1}{2}-\frac{1}{2}\cos 2x$sin²x = ½ − ½ cos 2x says: a constant ½, and a wave of size ½ about it.

Integrating the pieces

Now $\int {\sin }^{2}xdx=\int \left(\frac{1}{2}-\frac{1}{2}\cos 2x\right)dx$∫ sin²x dx = ∫ (½ − ½ cos 2x) dx. The ½ integrates to $\frac{x}{2}$x/2. The cos 2x has a 2 inside, so it integrates to $\frac{\sin 2x}{2}$(sin 2x)/2, and ½ of that is $\frac{\sin 2x}{4}$(sin 2x)/4. Together, $\int {\sin }^{2}xdx=\frac{x}{2}-\frac{\sin 2x}{4}+C$∫ sin²x dx = x/2 − (sin 2x)/4 + C.

Check by differentiating. $\frac{x}{2}$x/2 gives ½, and $\frac{\sin 2x}{4}$(sin 2x)/4 gives $\frac{2\cos 2x}{4}=\frac{1}{2}\cos 2x$(2 cos 2x)/4 = ½ cos 2x. So the derivative is ½ − ½ cos 2x, which is ${\sin }^{2}x$sin²x.

${\cos }^2$cos² takes a plus

Another form of the same formula is $\cos 2x=2{\cos }^{2}x-1$cos 2x = 2 cos²x − 1. Rearranged, ${\cos }^{2}x=\frac{1+\cos 2x}{2}$cos²x = (1 + cos 2x)/2: the same move, with a plus where ${\sin }^2$sin² had a minus. Check at x = 0: ${\cos }^{2}0=1$cos²0 = 1, and $\frac{1+1}{2}=1$(1 + 1)/2 = 1.

So $\int {\cos }^{2}xdx=\frac{x}{2}+\frac{\sin 2x}{4}+C$∫ cos²x dx = x/2 + (sin 2x)/4 + C. Adding the two results gives $\int \left({\sin }^{2}x+{\cos }^{2}x\right)dx=x+C$∫ (sin²x + cos²x) dx = x + C, which is right, since ${\sin }^{2}x+{\cos }^{2}x=1$sin²x + cos²x = 1.

The area under $y={\cos }^{2}x$y = cos²x from 0 to X, measured against the line y = ½. The term $\frac{X}{2}$X/2 is the rectangle under the line; the term $\frac{\sin 2X}{4}$(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 $\frac{\pi }{2}$π/2, $\pi$π, $\frac{3\pi }{2}$3π/2 or $2\pi$2π, where sin 2X = 0: the parts above and below the line cancel, and the area is exactly $\frac{X}{2}$X/2.

Definite integrals

Over half a turn the wave term vanishes. The integral of ${\sin }^{2}x$sin²x from 0 to $\pi$π is ${\left[\frac{x}{2}-\frac{\sin 2x}{4}\right]}_0^{\pi }=\frac{\pi }{2}-0$[x/2 − (sin 2x)/4]₀^π = π/2 − 0, since $\sin 2\pi =0$sin 2π = 0 and sin 0 = 0. The integral of ${\cos }^{2}x$cos²x over the same interval is also $\frac{\pi }{2}$π/2, and the two add to $\pi$π, the integral of 1.

Over a part of a wave it does not. The integral of ${\sin }^{2}x$sin²x from 0 to $\frac{\pi }{4}$π/4 is $\frac{\pi }{8}-\frac{\sin \left(\frac{\pi }{2}\right)}{4}=\frac{\pi }{8}-\frac{1}{4}\approx 0.143$π/8 − (sin(π/2))/4 = π/8 − 1/4 ≈ 0.143. That is less than half of the width $\frac{\pi }{4}\approx 0.785$π/4 ≈ 0.785, because ${\sin }^{2}x$sin²x stays below ½ until $x=\frac{\pi }{4}$x = π/4.

Other angles and products

With 3x in place of x, the double angle is 6x: ${\sin }^2\left(3x\right)=\frac{1-\cos 6x}{2}$sin²(3x) = (1 − cos 6x)/2, so $\int {\sin }^2\left(3x\right)dx=\frac{x}{2}-\frac{\sin 6x}{12}+C$∫ sin²(3x) dx = x/2 − (sin 6x)/12 + C. Check: the derivative is $\frac{1}{2}-\frac{6\cos 6x}{12}=\frac{1}{2}-\frac{1}{2}\cos 6x$½ − (6 cos 6x)/12 = ½ − ½ cos 6x.

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

The usual mistakes

Integrating the square as a power. $\frac{{\sin }^{3}x}{3}$sin³x/3 differentiates to ${\sin }^{2}x\cos x$sin²x cos x, not ${\sin }^{2}x$sin²x.

The plus and minus swapped. ${\sin }^{2}x=\frac{1-\cos 2x}{2}$sin²x = (1 − cos 2x)/2 has the minus. At x = 0, ${\sin }^{2}x$sin²x is 0, and only the minus version gives $\frac{1-1}{2}=0$(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 $\frac{x}{2}-\frac{\sin 2x}{4}+C$x/2 − (sin 2x)/4 + C, not $x-\frac{\sin 2x}{4}+C$x − (sin 2x)/4 + C.

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

Using ${\sin }^{2}x+{\cos }^{2}x=1$sin²x + cos²x = 1. It only trades ${\sin }^{2}x$sin²x for $1-{\cos }^{2}x$1 − cos²x, which still has a square.