Use a half-angle identity for a squared sine, integrals from easy to advanced lesson

The graph of $\sin^2 x$ on $[0, \pi]$ is a hump that starts at $0$ , rises to $1$ at $x = \pi/2$ and returns to $0$ at $x = \pi$ . How large is the area under it? Here $\sin^2 x$ means $(\sin x)^2$ , not $\sin(x^2)$ , and angles are in radians. The substitution $u = \sin x$ would need a factor $\cos x$ that is not there, so this page rewrites the square with the cosine of a doubled angle instead, integrates the two pieces that result, and shows why the exact area is $\pi/2$ .

Key terms

Cosine of a doubled angle

Definition For every real number $x$ ,

\begin{aligned} \cos 2x &= \cos^2 x - \sin^2 x \\[0.4em] &= 1 - 2\sin^2 x \\[0.4em] &= 2\cos^2 x - 1. \end{aligned}

The first form is the angle addition formula $\cos(a + b) = \cos a\cos b - \sin a\sin b$ with $a = b = x$ , and the other two follow from $\sin^2 x + \cos^2 x = 1$ . As with sine, $\cos^2 x$ means $(\cos x)^2$ .

In plain words Doubling the angle turns a difference of squares of sine and cosine into a single cosine. Read backwards, a square of sine or of cosine can be written with a cosine of twice the angle and no square at all.

Example At $x = \pi/6$ the left side is $\cos(\pi/3) = \tfrac12$ , and the second form gives $1 - 2\sin^2(\pi/6) = 1 - 2\cdot\tfrac14 = \tfrac12$ . As $x$ runs from $0$ to $\pi$ , the doubled angle $2x$ runs from $0$ to $2\pi$ , so $\cos 2x$ goes through one full period, from $1$ down to $-1$ at $x = \pi/2$ and back to $1$ .

Power-reducing identity

Definition For every real number $x$ ,

\begin{gathered} \sin^2 x = \frac{1 - \cos 2x}{2}, \\[0.4em] \cos^2 x = \frac{1 + \cos 2x}{2}. \end{gathered}

They come from solving $\cos 2x = 1 - 2\sin^2 x$ and $\cos 2x = 2\cos^2 x - 1$ for the square. Both are exact at every $x$ , not approximations, and they replace a second power of sine or cosine by a first power of cosine.

In plain words A squared sine is a constant height $\tfrac12$ plus a wave $-\tfrac12\cos 2x$ that swings between $-\tfrac12$ and $\tfrac12$ with period $\pi$ . The square is gone, and both pieces have known antiderivatives.

Example In the form $\sin^2 x = \tfrac12 - \tfrac12\cos 2x$ , at $x = 0$ the wave is $-\tfrac12$ and the total is $0$ . At $x = \pi/2$ the wave is $\tfrac12$ , because $\cos\pi = -1$ , and the total is $1$ , matching $\sin^2(\pi/2) = 1$ . The hump lies below the height $\tfrac12$ for $0 \le x < \pi/4$ and for $3\pi/4 < x \le \pi$ , and above it for $\pi/4 < x < 3\pi/4$ .

Half-angle identity

Definition For every real number $\theta$ ,

\begin{aligned} \sin\frac{\theta}{2} &= \pm\sqrt{\frac{1 - \cos\theta}{2}}, \\[0.4em] \cos\frac{\theta}{2} &= \pm\sqrt{\frac{1 + \cos\theta}{2}}, \end{aligned}

where each sign is the sign of the left side, which depends on the quadrant of $\theta/2$ . These are the power-reducing identities with $x = \theta/2$ , solved for the sine or the cosine itself.

In plain words The same identity, read the other way, finds the sine or cosine of half an angle from the cosine of the whole angle. A square root is never negative, so the sign has to come from knowing where the half angle lies.

Example With $\theta = \pi/4$ we have $\cos\theta = \sqrt2/2$ , and since $\pi/8$ lies in the first quadrant the sign is plus,

\begin{aligned} \sin\frac{\pi}{8} &= \sqrt{\frac{1 - \sqrt2/2}{2}} \\[0.4em] &= \frac{\sqrt{2 - \sqrt2}}{2} \approx 0.3827. \end{aligned}

With $\theta = 5\pi/4$ we have $\cos\theta = -\sqrt2/2$ , and the cosine formula gives the same root, $\sqrt{2 - \sqrt2}/2$ . But $5\pi/8$ lies in the second quadrant, where cosine is negative, so $\cos(5\pi/8) = -\sqrt{2 - \sqrt2}/2 \approx -0.3827$ .

Integral of sine squared

Definition With angles in radians, the power-reducing identity and linearity give, on the whole real line,

\begin{gathered} \int \sin^2 x\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C, \\[0.4em] \int \cos^2 x\,dx = \frac{x}{2} + \frac{\sin 2x}{4} + C. \end{gathered}

The $\tfrac14$ combines the $\tfrac12$ in front of $\cos 2x$ in the identity with the $\tfrac12$ in the antiderivative $\tfrac12\sin 2x$ of $\cos 2x$ , which compensates for the inner derivative $2$ . Radians matter because the derivative of $\sin x$ is $\cos x$ only when $x$ is measured in radians.

In plain words Split the square into the constant $\tfrac12$ and the wave $-\tfrac12\cos 2x$ , integrate each piece, and keep both factors of $\tfrac12$ in the coefficient of the wave.

Example For the hump on $[0, \pi]$ ,

\begin{aligned} \int_0^\pi \sin^2 x\,dx &= \Bigl[\frac{x}{2} - \frac{\sin 2x}{4}\Bigr]_0^\pi \\[0.4em] &= \frac{\pi}{2} - \frac{\sin 2\pi}{4} + \frac{\sin 0}{4} \\[0.4em] &= \frac{\pi}{2}. \end{aligned}

Both sine values are $0$ , so the wave contributes nothing and the area is $\pi/2 \approx 1.571$ , the same as a rectangle of height $\tfrac12$ and width $\pi$ . In other words, the average value of $\sin^2 x$ on $[0, \pi]$ is $\tfrac12$ . Differentiating gives $\tfrac12 - \tfrac12\cos 2x = \sin^2 x$ , which checks the coefficient, the sign and the doubled angle at once. Since $\sin^2 x \ge 0$ , the value $\pi/2$ is an ordinary area. The function $\sin(x^2)$ is a different integrand, and its integral over $[0, \pi]$ is about $0.773$ .

Cancellation over a full period

Definition Let $g$ be continuous on the real line and periodic with period $T > 0$ , so $g(x + T) = g(x)$ for every $x$ . Then the integral of $g$ over an interval of length $T$ does not depend on where the interval starts, $\int_a^{a+T} g(x)\,dx = \int_0^T g(x)\,dx$ for every real $a$ . For $g(x) = \cos(kx)$ with $k \ne 0$ the period is $2\pi/\lvert k \rvert$ and that common value is $0$ ,

\int_a^{a + 2\pi/\lvert k \rvert} \cos(kx)\,dx = 0.

The same holds for $\sin(kx)$ , and for any whole number of periods. Over an interval that is not a whole number of periods the integral may or may not be $0$ .

In plain words Over one full cycle a cosine wave encloses as much area above the axis as below it, so its signed contributions cancel. That is why the hump of $\sin^2 x$ on $[0, \pi]$ has the same area as the flat height $\tfrac12$ . The parts of the hump below $\tfrac12$ near the ends balance the part above $\tfrac12$ in the middle. The cancellation happens in the wave $-\tfrac12\cos 2x$ , which has signed parts, not in $\sin^2 x$ , which is never negative.

Example On $[0, \pi]$ the doubled angle makes exactly one period, and

\int_0^\pi \cos 2x\,dx = \Bigl[\frac{\sin 2x}{2}\Bigr]_0^\pi = 0.

On $[0, \pi/2]$ , half a period, the integral is also $0$ , because the positive part on $[0, \pi/4]$ and the negative part on $[\pi/4, \pi/2]$ happen to balance. On $[0, \pi/4]$ there is nothing to balance the positive part, and $\int_0^{\pi/4} \cos 2x\,dx = \tfrac12\sin(\pi/2) = \tfrac12$ . A full period is a reliable reason for cancellation, but it has to be checked for the bounds at hand.

Check yourself

Evaluate the integral

\int_0^{\pi/2} \cos^2 x\,dx.

Show the answer

Use $\cos^2 x = \tfrac12 + \tfrac12\cos 2x$ . Then

\begin{aligned} &\int_0^{\pi/2} \cos^2 x\,dx \\[0.4em] &\quad = \Bigl[\frac{x}{2} + \frac{\sin 2x}{4}\Bigr]_0^{\pi/2} \\[0.4em] &\quad = \frac{\pi}{4} + \frac{\sin\pi}{4} - 0 = \frac{\pi}{4}. \end{aligned}

The wave contributes nothing because $\sin\pi = \sin 0 = 0$ . As a check, $\sin^2 x + \cos^2 x = 1$ means the integrals of $\sin^2 x$ and $\cos^2 x$ over $[0, \pi/2]$ add to $\pi/2$ , and the same calculation gives $\pi/4$ for each.

Evaluate the integral below. A classmate answers $\pi/6$ , half the length of the interval. What did the classmate assume, and why does it fail here?

\int_0^{\pi/3} \sin^2 x\,dx

Show the answer

The antiderivative gives

\begin{aligned} \int_0^{\pi/3} \sin^2 x\,dx &= \Bigl[\frac{x}{2} - \frac{\sin 2x}{4}\Bigr]_0^{\pi/3} \\[0.4em] &= \frac{\pi}{6} - \frac{\sqrt3}{8} \approx 0.307, \end{aligned}

because $\sin(2\pi/3) = \sqrt3/2$ . The classmate assumed that the wave $-\tfrac12\cos 2x$ cancels. As $x$ runs from $0$ to $\pi/3$ , the doubled angle runs only from $0$ to $2\pi/3$ , a third of a period, and the wave leaves the contribution $-\sqrt3/8$ . In general $\int_0^b \sin^2 x\,dx = \tfrac{b}{2} - \tfrac14\sin 2b$ , which equals $b/2$ exactly when $b$ is a whole-number multiple of $\pi/2$ . The smaller value also fits the graph, since the hump stays below $\tfrac12$ on $[0, \pi/4]$ , three quarters of this interval.

Evaluate the integral below by applying the power-reducing identity twice. Is the answer the square of $\pi/2$ ?

\int_0^\pi \sin^4 x\,dx

Show the answer

Square the identity for $\sin^2 x$ , then use $\cos^2 2x = \tfrac12 + \tfrac12\cos 4x$ ,

\begin{aligned} \sin^4 x &= \Bigl(\frac{1 - \cos 2x}{2}\Bigr)^2 \\[0.4em] &= \frac14 - \frac{\cos 2x}{2} + \frac{\cos^2 2x}{4} \\[0.4em] &= \frac38 - \frac{\cos 2x}{2} + \frac{\cos 4x}{8}. \end{aligned}

On $[0, \pi]$ the function $\cos 2x$ makes one full period and $\cos 4x$ makes two, so both integrate to $0$ , and

\int_0^\pi \sin^4 x\,dx = \frac{3\pi}{8} \approx 1.178.

It is not $(\pi/2)^2 \approx 2.467$ . The integrand $\sin^4 x$ is the square of $\sin^2 x$ , but the integral of a square is not the square of the integral.

Use a half-angle identity for a squared sine

Key terms

Cosine of a doubled angle

Power-reducing identity

Half-angle identity

Integral of sine squared

Cancellation over a full period

Check yourself

How this lesson connects

Builds on

Used later in