Choosing a Trigonometric Substitution

When u-substitution has nothing to cancel

Take $\int \sqrt{4 - x^{2}} \,dx$. The substitution $u = 4 - x^{2}$ would need its derivative, −2x, beside the root to cancel du. There is no factor of x there, so that substitution leaves a mixture of u and x.

Compare $\int x\sqrt{4 - x^{2}} \,dx$, which does have the x: with $u = 4 - x^{2}$ it becomes $-\frac{1}{2} \int \sqrt{u} \,du = -\frac{1}{3}\left(4 - x^{2}\right)^{\frac{3}{2}} + C$. Without the x, a different substitution is needed.

Substituting $x = 2 \sin \theta$

Let $x = 2 \sin \theta$, with $\theta$ between $- \frac{\pi}{2}$ and $\frac{\pi}{2}$, so that x runs from −2 to 2 as $\theta$ runs across. Then $4 - x^{2} = 4 - 4\sin^{2}\theta = 4\left(1 - \sin^{2}\theta\right) = 4\cos^{2}\theta$, by the identity $\sin^{2}\theta + \cos^{2}\theta = 1$.

On that range $\cos \theta$ is not negative, so $\sqrt{4 - x^{2}} = 2 \cos \theta$. The root is gone.

A right triangle with hypotenuse a = 5 and angle $\theta$. The side opposite $\theta$ is $x = 5 \sin \theta$, and by Pythagoras the side next to it is $\sqrt{25 - x^{2}} = 5 \cos \theta$. At $\theta = 45^\circ$ both sides are 3.54. Drag $\theta$ to 53° and the side next to $\theta$ is 3.01 while x is 3.99.

The integral in $\theta$

The dx changes too. Differentiating $x = 2 \sin \theta$ gives $dx = 2 \cos \theta d\theta$. So $\int \sqrt{4 - x^{2}} \,dx = \int 2 \cos \theta \times 2 \cos \theta d\theta = \int 4\cos^{2}\theta d\theta$.

A cosine squared needs the double angle formula: $\cos^{2}\theta = \frac{{1 + \cos 2\theta}}{2}$. So $\int 4\cos^{2}\theta d\theta = \int 2\left(1 + \cos 2\theta\right) d\theta = 2\theta + \sin 2\theta + C$.

Back to x through a triangle

The answer has to be in x. From $x = 2 \sin \theta$, $\sin \theta = \frac{x}{2}$, so $\theta = \sin^{-1}\left(\frac{x}{2}\right)$.

For $\sin 2\theta = 2 \sin \theta \cos \theta$, $\cos \theta$ is needed as well. Draw a right triangle with angle $\theta$, hypotenuse 2 and opposite side x, so that $\sin \theta = \frac{x}{2}$. By Pythagoras the third side is $\sqrt{4 - x^{2}}$, so $\cos \theta = \frac{\sqrt{4 - x^{2}}}{2}$.

Then $\sin 2\theta = 2 \times \left(\frac{x}{2}\right) \times \frac{\sqrt{4 - x^{2}}}{2} = \frac{x\sqrt{4 - x^{2}}}{2}$, and $\int \sqrt{4 - x^{2}} \,dx = 2 \sin^{-1}\left(\frac{x}{2}\right) + \frac{x\sqrt{4 - x^{2}}}{2} + C$.

Changing the limits

In a definite integral the limits can travel with the substitution instead, and then there is no need to go back to x. For the integral of $\sqrt{4 - x^{2}}$ from 0 to 2: at x = 0, $\sin \theta = 0$, so $\theta = 0$, and at x = 2, $\sin \theta = 1$, so $\theta = \frac{\pi}{2}$. The value is $\left[2\theta + \sin 2\theta\right]_{0}^{\frac{\pi}{2}} = \left(\pi + \sin \pi\right) - 0 = \pi$.

That is right: $y = \sqrt{4 - x^{2}}$ is the top half of the circle of radius 2, and from 0 to 2 the region is a quarter of that circle, of area $\frac{1}{4} \times \pi \times 2^{2} = \pi$.

From 0 to 1, the upper limit becomes $\theta = \frac{\pi}{6}$, since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$. The value is $2\left(\frac{\pi}{6}\right) + \sin\left(\frac{\pi}{3}\right) = \frac{\pi}{3} + \frac{\sqrt{3}}{2} \approx 1.913$.

The top half of the circle $x^{2} + y^{2} = 4$, shaded from x = 0 to x = 1. The region is a triangle with base 1 and height $\sqrt{3}$, of area $\frac{\sqrt{3}}{2}$, plus a sector of the circle between the radii to $\left(1, \sqrt{3}\right)$ and (0, 2), whose angle is $\frac{\pi}{6}$ and whose area is $\frac{1}{2} \times 2^{2} \times \frac{\pi}{6} = \frac{\pi}{3}$. Together that is $\frac{\pi}{3} + \frac{\sqrt{3}}{2} \approx 1.913$, the value of the integral.

The sign inside decides

Each substitution rests on an identity that turns a sum or difference of squares into one square. $a^{2} - x^{2}$ takes $x = a \sin \theta$, because $1 - \sin^{2}\theta = \cos^{2}\theta$. $a^{2} + x^{2}$ takes $x = a \tan \theta$, because $1 + \tan^{2}\theta = \sec^{2}\theta$. $x^{2} - a^{2}$ takes $x = a \sec \theta$, because $\sec^{2}\theta - 1 = \tan^{2}\theta$.

The wrong choice leaves a sum that is not a square: with $x = 5 \tan \theta$, $25 - x^{2}$ becomes $25 - 25\tan^{2}\theta$, which no identity simplifies.

A plus inside

For $\int \frac{1}{{9 + x^{2}}} \,dx$, let $x = 3 \tan \theta$. Then $dx = 3\sec^{2}\theta d\theta$, and $9 + x^{2} = 9 + 9\tan^{2}\theta = 9\sec^{2}\theta$. The integral becomes $\int \frac{3\sec^{2}\theta}{{9\sec^{2}\theta}} d\theta = \int \frac{1}{3} d\theta = \frac{\theta}{3} + C$. Since $\theta = \tan^{-1}\left(\frac{x}{3}\right)$, that is $\left(\frac{1}{3}\right) \tan^{-1}\left(\frac{x}{3}\right) + C$.

The triangle does the work for a tangent substitution too. For $\int \frac{1}{\left(1 + x^{2}\right)^{\frac{3}{2}}} \,dx$, let $x = \tan \theta$, so $dx = \sec^{2}\theta d\theta$ and $\left(1 + x^{2}\right)^{\frac{3}{2}} = \sec^{3}\theta$. The integral becomes $\int \frac{\sec^{2}\theta}{\sec^{3}\theta} d\theta = \int \cos \theta d\theta = \sin \theta + C$.

To go back, draw a right triangle with $\tan \theta = x$: opposite side x, adjacent side 1, hypotenuse $\sqrt{1 + x^{2}}$. Then $\sin \theta = \frac{x}{\sqrt{1 + x^{2}}}$, and the integral is $\frac{x}{\sqrt{1 + x^{2}}} + C$. From 0 to 1 it is $\frac{1}{\sqrt{2}} \approx 0.707$.

The usual mistakes

Leaving out dx. Replacing $\sqrt{4 - x^{2}}$ by $2 \cos \theta$ and stopping gives $\int 2 \cos \theta d\theta$, which is missing a whole factor of $2 \cos \theta$; $dx = 2 \cos \theta d\theta$ supplies it.

Keeping the old limits. After $x = 2 \sin \theta$ the variable is an angle, so the limits 0 and 2 become 0 and $\frac{\pi}{2}$.

Leaving the answer in $\theta$. An indefinite integral in x needs an answer in x, which the right triangle gives.

Tangent for a minus, or sine for a plus. The identity has to make a single square: $1 - \sin^{2}\theta$ for $a^{2} - x^{2}$, $1 + \tan^{2}\theta$ for $a^{2} + x^{2}$.