Choosing a Trigonometric Substitution

When u-substitution has nothing to cancel

Take $\int \sqrt{4-x^2}dx$∫ √(4 − x²) dx. The substitution $u=4-x^2$u = 4 − x² 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 $\int x\sqrt{4-x^2}dx$∫ x√(4 − x²) dx, which does have the x: with $u=4-x^2$u = 4 − x² it becomes $-\frac{1}{2}\int \sqrt{u}du=-\frac{1}{3}{\left(4-x^2\right)}^{\frac{3}{2}}+C$−½ ∫ √u du = −⅓(4 − x²)^(3/2) + C. Without the x, a different substitution is needed.

Substituting $x=2\sin \theta$x = 2 sin θ

Let $x=2\sin \theta$x = 2 sin θ, with $\theta$θ between $-\frac{\pi }{2}$−π/2 and $\frac{\pi }{2}$π/2, so that x runs from −2 to 2 as $\theta$θ runs across. Then $4-x^2=4-4{\sin }^2\theta =4\left(1-{\sin }^2\theta \right)=4{\cos }^2\theta$4 − x² = 4 − 4sin²θ = 4(1 − sin²θ) = 4cos²θ, by the identity ${\sin }^2\theta +{\cos }^2\theta =1$sin²θ + cos²θ = 1.

On that range $\cos \theta$cos θ is not negative, so $\sqrt{4-x^2}=2\cos \theta$√(4 − x²) = 2 cos θ. The root is gone.

A right triangle with hypotenuse a = 5 and angle $\theta$θ. The side opposite $\theta$θ is $x=5\sin \theta$x = 5 sin θ, and by Pythagoras the side next to it is $\sqrt{25-x^2}=5\cos \theta$√(25 − x²) = 5 cos θ. At $\theta =45^{\circ }$θ = 45° both sides are 3.54. Drag $\theta$θ to 53° and the side next to $\theta$θ is 3.01 while x is 3.99.

The integral in $\theta$θ

The dx changes too. Differentiating $x=2\sin \theta$x = 2 sin θ gives $dx=2\cos \theta d\theta$dx = 2 cos θ dθ. So $\int \sqrt{4-x^2}dx=\int 2\cos \theta \times 2\cos \theta d\theta =\int 4{\cos }^2\theta d\theta$∫ √(4 − x²) dx = ∫ 2 cos θ × 2 cos θ dθ = ∫ 4cos²θ dθ.

A cosine squared needs the double angle formula: ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$cos²θ = (1 + cos 2θ)/2. So $\int 4{\cos }^2\theta d\theta =\int 2\left(1+\cos 2\theta \right)d\theta =2\theta +\sin 2\theta +C$∫ 4cos²θ dθ = ∫ 2(1 + cos 2θ) dθ = 2θ + sin 2θ + C.

Back to x through a triangle

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

For $\sin 2\theta =2\sin \theta \cos \theta$sin 2θ = 2 sin θ cos θ, $\cos \theta$cos θ is needed as well. Draw a right triangle with angle $\theta$θ, hypotenuse 2 and opposite side x, so that $\sin \theta =\frac{x}{2}$sin θ = x/2. By Pythagoras the third side is $\sqrt{4-x^2}$√(4 − x²), so $\cos \theta =\frac{\sqrt{4-x^2}}{2}$cos θ = √(4 − x²)/2.

Then $\sin 2\theta =2\times \left(\frac{x}{2}\right)\times \frac{\sqrt{4-x^2}}{2}=\frac{x\sqrt{4-x^2}}{2}$sin 2θ = 2 × (x/2) × √(4 − x²)/2 = x√(4 − x²)/2, and $\int \sqrt{4-x^2}dx=2{\sin }^{-1}\left(\frac{x}{2}\right)+\frac{x\sqrt{4-x^2}}{2}+C$∫ √(4 − x²) dx = 2 sin⁻¹(x/2) + x√(4 − x²)/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 $\sqrt{4-x^2}$√(4 − x²) from 0 to 2: at x = 0, $\sin \theta =0$sin θ = 0, so $\theta =0$θ = 0, and at x = 2, $\sin \theta =1$sin θ = 1, so $\theta =\frac{\pi }{2}$θ = π/2. The value is ${\left[2\theta +\sin 2\theta \right]}_0^{\frac{\pi }{2}}=\left(\pi +\sin \pi \right)-0=\pi$[2θ + sin 2θ]₀^(π/2) = (π + sin π) − 0 = π.

That is right: $y=\sqrt{4-x^2}$y = √(4 − x²) 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 $\frac{1}{4}\times \pi \times 2^2=\pi$¼ × π × 2² = π.

From 0 to 1, the upper limit becomes $\theta =\frac{\pi }{6}$θ = π/6, since $\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}$sin(π/6) = ½. 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$2(π/6) + sin(π/3) = π/3 + √3/2 ≈ 1.913.

The top half of the circle $x^2+y^2=4$x² + y² = 4, shaded from x = 0 to x = 1. The region is a triangle with base 1 and height $\sqrt{3}$√3, of area $\frac{\sqrt{3}}{2}$√3/2, plus a sector of the circle between the radii to $\left(1,\sqrt{3}\right)$(1, √3) and (0, 2), whose angle is $\frac{\pi }{6}$π/6 and whose area is $\frac{1}{2}\times 2^2\times \frac{\pi }{6}=\frac{\pi }{3}$½ × 2² × π/6 = π/3. Together that is $\frac{\pi }{3}+\frac{\sqrt{3}}{2}\approx 1.913$π/3 + √3/2 ≈ 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. $a^2-x^2$a² − x² takes $x=a\sin \theta$x = a sin θ, because $1-{\sin }^2\theta ={\cos }^2\theta$1 − sin²θ = cos²θ. $a^2+x^2$a² + x² takes $x=a\tan \theta$x = a tan θ, because $1+{\tan }^2\theta ={\sec }^2\theta$1 + tan²θ = sec²θ. $x^2-a^2$x² − a² takes $x=a\sec \theta$x = a sec θ, because ${\sec }^2\theta -1={\tan }^2\theta$sec²θ − 1 = tan²θ.

The wrong choice leaves a sum that is not a square: with $x=5\tan \theta$x = 5 tan θ, $25-x^2$25 − x² becomes $25-25{\tan }^2\theta$25 − 25tan²θ, which no identity simplifies.

A plus inside

For $\int \frac{1}{9+x^2}dx$∫ 1/(9 + x²) dx, let $x=3\tan \theta$x = 3 tan θ. Then $dx=3{\sec }^2\theta d\theta$dx = 3sec²θ dθ, and $9+x^2=9+9{\tan }^2\theta =9{\sec }^2\theta$9 + x² = 9 + 9tan²θ = 9sec²θ. The integral becomes $\int \frac{3{\sec }^2\theta }{9{\sec }^2\theta }d\theta =\int \frac{1}{3}d\theta =\frac{\theta }{3}+C$∫ 3sec²θ/(9sec²θ) dθ = ∫ ⅓ dθ = θ/3 + C. Since $\theta ={\tan }^{-1}\left(\frac{x}{3}\right)$θ = tan⁻¹(x/3), that is $\left(\frac{1}{3}\right){\tan }^{-1}\left(\frac{x}{3}\right)+C$(1/3) tan⁻¹(x/3) + C.

The triangle does the work for a tangent substitution too. For $\int \frac{1}{{\left(1+x^2\right)}^{\frac{3}{2}}}dx$∫ 1/(1 + x²)^(3/2) dx, let $x=\tan \theta$x = tan θ, so $dx={\sec }^2\theta d\theta$dx = sec²θ dθ and ${\left(1+x^2\right)}^{\frac{3}{2}}={\sec }^3\theta$(1 + x²)^(3/2) = sec³θ. The integral becomes $\int \frac{{\sec }^2\theta }{{\sec }^3\theta }d\theta =\int \cos \theta d\theta =\sin \theta +C$∫ sec²θ/sec³θ dθ = ∫ cos θ dθ = sin θ + C.

To go back, draw a right triangle with $\tan \theta =x$tan θ = x: opposite side x, adjacent side 1, hypotenuse $\sqrt{1+x^2}$√(1 + x²). Then $\sin \theta =\frac{x}{\sqrt{1+x^2}}$sin θ = x/√(1 + x²), and the integral is $\frac{x}{\sqrt{1+x^2}}+C$x/√(1 + x²) + C. From 0 to 1 it is $\frac{1}{\sqrt{2}}\approx 0.707$1/√2 ≈ 0.707.

The usual mistakes

Leaving out dx. Replacing $\sqrt{4-x^2}$√(4 − x²) by $2\cos \theta$2 cos θ and stopping gives $\int 2\cos \theta d\theta$∫ 2 cos θ dθ, which is missing a whole factor of $2\cos \theta$2 cos θ; $dx=2\cos \theta d\theta$dx = 2 cos θ dθ supplies it.

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

Leaving the answer in $\theta$θ. 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$1 − sin²θ for $a^2-x^2$a² − x², $1+{\tan }^2\theta$1 + tan²θ for $a^2+x^2$a² + x².