Trigonometric substitution for a circle inside a square root, integrals from easy to advanced lesson

The integral of $1/\sqrt{4 - x^2}$ from $0$ to $1$ has a square root in the denominator and no matching inner derivative in the numerator. The expression under the root is $2^2 - x^2$ , a squared radius minus $x^2$ , and its square root is the height of a circle of radius $2$ above the point $x$ . Putting $x = 2\sin\theta$ makes the root equal to $2\cos\theta$ and the differential equal to $2\cos\theta\,d\theta$ , so the integrand reduces to $1$ in the angle variable. This page explains that trigonometric substitution, how to choose angle bounds on which the square root keeps its sign, and how the inverse sine confirms the value $\pi/6$ .

Key terms

Trigonometric substitution

Definition Trigonometric substitution is the substitution rule with $x$ written as a trigonometric function of a new variable $\theta$ . The rule says that if $g$ has a continuous derivative on $[\alpha, \beta]$ and $f$ is continuous on an interval containing every value $g(\theta)$ with $\theta$ in $[\alpha, \beta]$ , then

\begin{aligned} &\int_{g(\alpha)}^{g(\beta)} f(x)\,dx \\[0.4em] &\quad = \int_{\alpha}^{\beta} f(g(\theta))\,g'(\theta)\,d\theta. \end{aligned}

For a constant $a > 0$ , three standard choices clear a square root with a Pythagorean identity, each on an angle range where the root has a known sign. For $\sqrt{a^2 - x^2}$ with $-a \le x \le a$ , take $x = a\sin\theta$ with $-\pi/2 \le \theta \le \pi/2$ , where $\cos\theta \ge 0$ and the root equals $a\cos\theta$ . For $\sqrt{a^2 + x^2}$ , take $x = a\tan\theta$ with $-\pi/2 < \theta < \pi/2$ , where $\sec\theta > 0$ and the root equals $a\sec\theta$ . For $\sqrt{x^2 - a^2}$ with $x \ge a$ , take $x = a\sec\theta$ with $0 \le \theta < \pi/2$ , where $\tan\theta \ge 0$ and the root equals $a\tan\theta$ . When $x \le -a$ , the range $\pi/2 < \theta \le \pi$ is used instead, and there the root equals $-a\tan\theta$ . On each of these ranges the substitution is one-to-one, so $\theta$ can be recovered from $x$ when an antiderivative has to be written back in $x$ .

In plain words A constant square plus or minus $x^2$ under a root has the shape of a Pythagorean identity. Choose $x$ as a sine, tangent or secant so that the expression under the root becomes a single square, keep the angle in a range where the root of that square has a known sign, and change $dx$ and the bounds along with the integrand.

Example For the integral of this page, the integrand is continuous on $[0, 1]$ , since $4 - x^2 \ge 3$ there. Take $x = 2\sin\theta$ . As $\theta$ runs from $0$ to $\pi/6$ , the value of $x$ runs from $0$ to $1$ , the differential is $dx = 2\cos\theta\,d\theta$ , and $\cos\theta \ge \sqrt{3}/2 > 0$ , so $\sqrt{4 - 4\sin^2\theta} = 2\cos\theta$ . Hence

\begin{aligned} &\int_0^1 \frac{dx}{\sqrt{4 - x^2}} \\[0.4em] &\quad = \int_0^{\pi/6} \frac{2\cos\theta}{2\cos\theta}\,d\theta \\[0.4em] &\quad = \int_0^{\pi/6} 1\,d\theta = \frac{\pi}{6}, \end{aligned}

which is about $0.5236$ . The factor $2\cos\theta$ is never $0$ on $[0, \pi/6]$ , so the cancellation is valid. The circle identity simplified the denominator, and the derivative of $2\sin\theta$ supplied the same factor in the numerator.

Sine substitution

Definition Let $a > 0$ . The sine substitution is $x = a\sin\theta$ with $-\pi/2 \le \theta \le \pi/2$ . On this interval the sine increases from $-1$ to $1$ , so each $x$ in $[-a, a]$ comes from exactly one angle, $\theta = \arcsin(x/a)$ . Since $1 - \sin^2\theta = \cos^2\theta$ and $\cos\theta \ge 0$ on $[-\pi/2, \pi/2]$ , the substitution gives

\begin{gathered} a^2 - x^2 = a^2\cos^2\theta, \\[0.4em] \sqrt{a^2 - x^2} = a\cos\theta, \\[0.4em] dx = a\cos\theta\,d\theta. \end{gathered}

In plain words The root and the differential produce the same factor $a\cos\theta$ . That is why $dx/\sqrt{a^2 - x^2}$ reduces to $d\theta$ , while $\sqrt{a^2 - x^2}\,dx$ becomes $a^2\cos^2\theta\,d\theta$ , which the half-angle identity integrates.

Example The integrand $\sqrt{9 - x^2}$ is continuous on $[0, 3]$ . With $x = 3\sin\theta$ , the bounds $0$ and $3$ become $0$ and $\pi/2$ , the root becomes $3\cos\theta$ , and the identity $\cos^2\theta = \tfrac12(1 + \cos 2\theta)$ gives

\begin{aligned} &\int_0^3 \sqrt{9 - x^2}\,dx \\[0.4em] &\quad = \int_0^{\pi/2} 9\cos^2\theta\,d\theta \\[0.4em] &\quad = \frac{9}{2}\Bigl[\theta + \tfrac12\sin 2\theta\Bigr]_0^{\pi/2} \\[0.4em] &\quad = \frac{9\pi}{4} \approx 7.0686. \end{aligned}

The graph of $\sqrt{9 - x^2}$ is the upper half of the circle of radius $3$ , and $9\pi/4$ is the area of a quarter of that disk, as it should be. At $\theta = \pi/2$ the root is $0$ , which is still nonnegative, so the endpoint causes no trouble.

Circle parameterization

Definition Let $a > 0$ . On the circle $x^2 + y^2 = a^2$ , measure the angle $\theta$ from the positive $y$ -axis, turning toward the positive $x$ -axis. The point at angle $\theta$ is

(x, y) = (a\sin\theta,\ a\cos\theta).

As $\theta$ runs over $[-\pi/2, \pi/2]$ , the point runs once along the upper half of the circle, from $(-a, 0)$ through $(0, a)$ to $(a, 0)$ , and its height is $y = a\cos\theta = \sqrt{a^2 - x^2}$ . The more familiar parameterization $(a\cos t, a\sin t)$ measures $t$ from the positive $x$ -axis. It gives the same points with $t = \pi/2 - \theta$ , and it swaps the roles of sine and cosine.

In plain words The substitution $x = a\sin\theta$ says that $x$ is the horizontal coordinate of a point on a circle of radius $a$ . The root $\sqrt{a^2 - x^2}$ is then the height of that point, and the identity $a^2 - a^2\sin^2\theta = a^2\cos^2\theta$ is the Pythagorean theorem in the right triangle with hypotenuse $a$ . The circle helps choose the variable. It need not be the graph whose area is computed.

Example For $a = 2$ , the point with $x = 1$ has $\sin\theta = 1/2$ , so $\theta = \pi/6$ , and its height is $2\cos(\pi/6) = \sqrt{3}$ . In the integral of this page the integrand $1/\sqrt{4 - x^2}$ equals $1/y$ , the reciprocal of the height of the circle, so the region being measured is not the region under the semicircle. The value $\pi/6$ equals the angle swept from $\theta = 0$ to $\theta = \pi/6$ , in radians. It is not the area of the circular sector with that angle, which is $\tfrac12 \cdot 2^2 \cdot \tfrac{\pi}{6} = \tfrac{\pi}{3}$ .

Angle bounds

Definition When a definite integral in $x$ is rewritten with $x = g(\theta)$ , the new bounds are angles $\alpha$ and $\beta$ such that $g(\alpha)$ is the old lower bound and $g(\beta)$ is the old upper bound, chosen inside the range on which the substitution is one-to-one and the root has a known sign. For the sine substitution $x = a\sin\theta$ and old bounds $p \le q$ in $[-a, a]$ , they are

\begin{gathered} \alpha = \arcsin\frac{p}{a}, \\[0.4em] \beta = \arcsin\frac{q}{a}, \end{gathered}

both in $[-\pi/2, \pi/2]$ , and as $\theta$ increases from $\alpha$ to $\beta$ , the value of $x$ increases from $p$ to $q$ . The angles are measured in radians.

In plain words Change the bounds together with the variable, and there is no need to substitute back at the end. Finding each new bound with the inverse sine puts it in the allowed range automatically. Another angle with the same sine gives the same endpoint, but the path between the bounds would then leave the range where the root equals $a\cos\theta$ .

Example For the integral of $1/\sqrt{4 - x^2}$ from $-1$ to $1$ , the new bounds are $\arcsin(-1/2) = -\pi/6$ and $\arcsin(1/2) = \pi/6$ . A negative angle is allowed, and $\cos\theta > 0$ on $[-\pi/6, \pi/6]$ , so

\begin{aligned} &\int_{-1}^{1} \frac{dx}{\sqrt{4 - x^2}} \\[0.4em] &\quad = \int_{-\pi/6}^{\pi/6} 1\,d\theta = \frac{\pi}{3}. \end{aligned}

The bounds must be radians. The angle $\pi/6$ is $30$ degrees, and bounds of $-30$ and $30$ in the $\theta$ integral would give a value about $57$ times too large.

Nonnegative square root

Definition For a number $s \ge 0$ , the symbol $\sqrt{s}$ denotes the nonnegative number whose square is $s$ . Consequently $\sqrt{t^2} = \lvert t \rvert$ for every real $t$ , and for $a > 0$

\sqrt{a^2\cos^2\theta} = a\lvert\cos\theta\rvert,

which equals $a\cos\theta$ exactly when $\cos\theta \ge 0$ , in particular for every $\theta$ in $[-\pi/2, \pi/2]$ .

In plain words Replacing $\sqrt{a^2\cos^2\theta}$ by $a\cos\theta$ is a claim about the sign of the cosine. It is true on the angle range you chose and false wherever the cosine is negative, so check the range before removing the absolute value.

Example On $[0, \pi/6]$ the cosine lies between $\sqrt{3}/2$ and $1$ , so $\sqrt{4\cos^2\theta} = 2\cos\theta$ there. At $\theta = 2\pi/3$ the cosine is $-1/2$ , and $\sqrt{4\cos^2\theta} = 1$ while $2\cos\theta = -1$ . A substitution that let $\theta$ go past $\pi/2$ would need the root written as $-2\cos\theta$ wherever the cosine is negative.

Inverse sine

Definition For $-1 \le y \le 1$ , the inverse sine $\arcsin y$ , also written $\sin^{-1} y$ , is the unique angle $\theta$ in $[-\pi/2, \pi/2]$ with $\sin\theta = y$ . It is the inverse function of the sine restricted to $[-\pi/2, \pi/2]$ , not the reciprocal $1/\sin y$ . For $-1 < y < 1$ it is differentiable, with

\frac{d}{dy}\arcsin y = \frac{1}{\sqrt{1 - y^2}},

because differentiating $\sin\theta = y$ gives $\cos\theta \cdot d\theta/dy = 1$ , and $\cos\theta = \sqrt{1 - y^2}$ since $\cos\theta > 0$ for $-\pi/2 < \theta < \pi/2$ . Hence for $a > 0$ , on the interval $-a < x < a$ ,

\begin{aligned} &\int \frac{dx}{\sqrt{a^2 - x^2}} \\[0.4em] &\quad = \arcsin\frac{x}{a} + C. \end{aligned}

In plain words The inverse sine reads the angle off from its sine. In a sine substitution, $\arcsin(x/a)$ is exactly the angle $\theta$ , so the antiderivative $\theta + C$ of the reduced integrand becomes $\arcsin(x/a) + C$ when it is written back in $x$ .

Example Let $F(x) = \arcsin(x/2)$ for $-2 < x < 2$ . The chain rule and the nonnegative root $\sqrt{1 - x^2/4} = \tfrac12\sqrt{4 - x^2}$ give

\begin{aligned} F'(x) &= \frac{1/2}{\sqrt{1 - x^2/4}} \\[0.4em] &= \frac{1}{\sqrt{4 - x^2}}. \end{aligned}

The integrand is continuous on $[0, 1]$ , so the fundamental theorem gives $F(1) - F(0) = \pi/6 - 0 = \pi/6$ , the value found above. A rough bound agrees. On $[0, 1]$ the integrand increases from $1/2$ to $1/\sqrt{3} \approx 0.5774$ , so over this interval of length $1$ the integral lies between $0.5$ and $0.5774$ , and $\pi/6 \approx 0.5236$ does.

Check yourself

Evaluate the integral below with a sine substitution, and check the result with the inverse sine.

\int_0^{\sqrt{2}} \frac{dx}{\sqrt{4 - x^2}}

Show the answer

The integrand is continuous on $[0, \sqrt{2}]$ , since $4 - x^2 \ge 2$ there. Put $x = 2\sin\theta$ . The bound $x = 0$ gives $\theta = 0$ , and $x = \sqrt{2}$ gives $\sin\theta = \sqrt{2}/2$ , so $\theta = \pi/4$ . On $[0, \pi/4]$ the cosine is at least $\sqrt{2}/2 > 0$ , so the root is $2\cos\theta$ , and with $dx = 2\cos\theta\,d\theta$

\begin{aligned} &\int_0^{\sqrt{2}} \frac{dx}{\sqrt{4 - x^2}} \\[0.4em] &\quad = \int_0^{\pi/4} 1\,d\theta = \frac{\pi}{4}. \end{aligned}

The inverse sine gives the same value, $\arcsin(\sqrt{2}/2) - \arcsin 0 = \pi/4 \approx 0.7854$ . As a rough check, the integrand rises from $1/2$ to $1/\sqrt{2}$ over an interval of length $\sqrt{2}$ , so the integral lies between $\sqrt{2}/2 \approx 0.7071$ and $1$ .

Evaluate the integral below, and explain the answer with the geometry of the circle.

\int_0^1 \sqrt{4 - x^2}\,dx

Show the answer

With $x = 2\sin\theta$ and $\theta$ from $0$ to $\pi/6$ , the root is $2\cos\theta$ and $dx = 2\cos\theta\,d\theta$ , so the new integrand is $4\cos^2\theta = 2 + 2\cos 2\theta$ . Then

\begin{aligned} &\int_0^1 \sqrt{4 - x^2}\,dx \\[0.4em] &\quad = \int_0^{\pi/6} (2 + 2\cos 2\theta)\,d\theta \\[0.4em] &\quad = \Bigl[2\theta + \sin 2\theta\Bigr]_0^{\pi/6} \\[0.4em] &\quad = \frac{\pi}{3} + \frac{\sqrt{3}}{2} \approx 1.9132. \end{aligned}

This time the graph is the upper half of the circle of radius $2$ , so the answer is an area in the circle picture. The radius to the point $(1, \sqrt{3})$ splits the region into a sector with angle $\pi/6$ and area $\pi/3$ , and a right triangle with legs $1$ and $\sqrt{3}$ and area $\sqrt{3}/2$ . The integrand decreases from $2$ to $\sqrt{3} \approx 1.732$ , and the value lies between those two numbers, as it must over an interval of length $1$ .

In the integral of this page, someone takes the upper bound $\theta = 5\pi/6$ , since $\sin(5\pi/6) = 1/2$ as well, replaces the root by $2\cos\theta$ , and gets $5\pi/6 \approx 2.618$ . What went wrong?

Show the answer

The angle $5\pi/6$ lies outside $[-\pi/2, \pi/2]$ , the range on which the sine substitution is one-to-one and the cosine is nonnegative, and two things fail. For $\pi/2 < \theta \le 5\pi/6$ the cosine is negative, so there $\sqrt{4\cos^2\theta} = -2\cos\theta$ , not $2\cos\theta$ . Also, as $\theta$ runs from $0$ to $5\pi/6$ , the point $x = 2\sin\theta$ climbs from $0$ to $2$ and comes back to $1$ , so it passes through $x = 2$ , where the integrand is undefined, and the substitution rule does not apply as stated. The inverse sine gives the correct bound $\arcsin(1/2) = \pi/6$ and the value $\pi/6$ . The rough bound catches the error as well, since the integral lies between $0.5$ and $0.5774$ .

Trigonometric substitution for a circle inside a square root

Key terms

Trigonometric substitution

Sine substitution

Circle parameterization

Angle bounds

Nonnegative square root

Inverse sine

Check yourself

How this lesson connects

Builds on