E6 - Trigonometric substitutions

E6 - Trigonometric substitutions

A trigonometric substitution can turn a square root into a familiar trigonometric factor. You will choose sine, tangent or secant by matching an identity, transform differentials and limits, and use a deliberate angle range to control the sign of a root.

Choosing sine, tangent or secant

How can you find the exact area ∫014−x2 dx\displaystyle\int_0^1\sqrt{4-x^2}\,\mathrm{d}x? The graph is part of a circle of radius 2, but the boundary at x=1x=1 cuts off only part of its first quadrant. A substitution will replace that awkward square root by a cosine.

Try x=2sin⁡θx=2\sin\theta. Then 4−x2=4(1−sin⁡2θ)=4cos⁡2θ4-x^2=4(1-\sin^2\theta)=4\cos^2\theta. Choosing −π/2≤θ≤π/2-\pi/2\le\theta\le\pi/2 makes cos⁡θ≥0\cos\theta\ge0, so the nonnegative square root is 2cos⁡θ2\cos\theta.

The same idea works with the other Pythagorean identities. Choose a>0a>0 and match the whole expression under the root.

ExpressionSubstitutionIdentity usedConvenient angle range
a2−x2\sqrt{a^2-x^2}x=asin⁡θx=a\sin\theta1−sin⁡2θ=cos⁡2θ1-\sin^2\theta=\cos^2\theta[−π/2,π/2][-\pi/2,\pi/2]
a2+x2\sqrt{a^2+x^2}x=atan⁡θx=a\tan\theta1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta(−π/2,π/2)(-\pi/2,\pi/2)
x2−a2\sqrt{x^2-a^2}x=asec⁡θx=a\sec\thetasec⁡2θ−1=tan⁡2θ\sec^2\theta-1=\tan^2\theta(0,π/2)(0,\pi/2) for x>ax>a

The secant range in the table covers positive xx. Negative xx needs a different branch, which we will handle in the last part. Exclude any endpoint where the original integrand is undefined.

Trigonometric substitution

A trigonometric substitution replaces an algebraic variable by a trigonometric expression in a new angle. Change the differential too: for example, x=asin⁡θx=a\sin\theta gives dx=acos⁡θ dθ\mathrm{d}x=a\cos\theta\,\mathrm{d}\theta. Every remaining occurrence of xx must be replaced.

Here are two simple demonstrations. Keep the numerator and scale fixed, and change only the sign inside the square root. A direct substitution based on the expression under the root would also work; these examples isolate why the trigonometric choices work.

Worked example 1: a sine substitution

Question. Use a trigonometric substitution to find ∫x4−x2 dx\displaystyle\int\frac{x}{\sqrt{4-x^2}}\,\mathrm{d}x, for −2<x<2-2<x<2.

Plan. Put x=2sin⁡θx=2\sin\theta, with −π/2<θ<π/2-\pi/2<\theta<\pi/2, so the root and differential contain the same positive cosine factor.

4−x2=2cos⁡θ,dx=2cos⁡θ dθ∫x4−x2 dx=∫2sin⁡θ2cos⁡θ 2cos⁡θ dθreplace every factor=2∫sin⁡θ dθcancel the cosines=−2cos⁡θ+C=−4−x2+Creturn to x\begin{aligned} \sqrt{4-x^2}&=2\cos\theta,\qquad \mathrm{d}x=2\cos\theta\,\mathrm{d}\theta\\ \int\frac{x}{\sqrt{4-x^2}}\,\mathrm{d}x &=\int\frac{2\sin\theta}{2\cos\theta}\,2\cos\theta\,\mathrm{d}\theta && \text{replace every factor}\\ &=2\int\sin\theta\,\mathrm{d}\theta && \text{cancel the cosines}\\ &=-2\cos\theta+C\\ &=-\sqrt{4-x^2}+C && \text{return to }x \end{aligned}

Answer. −4−x2+C-\sqrt{4-x^2}+C, on (−2,2)(-2,2).

Check. Differentiation gives −12(4−x2)−1/2(−2x)=x/4−x2-\frac12(4-x^2)^{-1/2}(-2x)=x/\sqrt{4-x^2}.

The plus sign selects tangent instead: 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta.

Worked example 2: change the sign, change the substitution

Question. Use a trigonometric substitution to find ∫x4+x2 dx\displaystyle\int\frac{x}{\sqrt{4+x^2}}\,\mathrm{d}x.

Plan. Put x=2tan⁡θx=2\tan\theta, with −π/2<θ<π/2-\pi/2<\theta<\pi/2, where sec⁡θ>0\sec\theta>0.

4+x2=2sec⁡θ,dx=2sec⁡2θ dθ∫x4+x2 dx=∫2tan⁡θ2sec⁡θ 2sec⁡2θ dθsubstitute throughout=2∫sec⁡θtan⁡θ dθcancel one secant factor=2sec⁡θ+C(sec⁡θ)′=sec⁡θtan⁡θ=4+x2+Csec⁡θ>0\begin{aligned} \sqrt{4+x^2}&=2\sec\theta,\qquad \mathrm{d}x=2\sec^2\theta\,\mathrm{d}\theta\\ \int\frac{x}{\sqrt{4+x^2}}\,\mathrm{d}x &=\int\frac{2\tan\theta}{2\sec\theta}\,2\sec^2\theta\,\mathrm{d}\theta && \text{substitute throughout}\\ &=2\int\sec\theta\tan\theta\,\mathrm{d}\theta && \text{cancel one secant factor}\\ &=2\sec\theta+C && (\sec\theta)'=\sec\theta\tan\theta\\ &=\sqrt{4+x^2}+C && \sec\theta>0 \end{aligned}

Answer. 4+x2+C\sqrt{4+x^2}+C, for all real xx.

Check. Differentiating the answer gives 12(4+x2)−1/2(2x)=x/4+x2\frac12(4+x^2)^{-1/2}(2x)=x/\sqrt{4+x^2}.

The substitutions are selected from identities; they are not arbitrary guesses. The derivative of secant is in the formula booklet, page 5. The Pythagorean identities and the three-choice substitution table are not printed there: know the identities and use them to reconstruct the table. The angle-addition identities on page 4 let you derive the double-angle identity used next.

Watch out

Wrong: x=2tan⁡θx=2\tan\theta, then dx=2sec⁡θ dθ\mathrm{d}x=2\sec\theta\,\mathrm{d}\theta. Differentiate tangent correctly: dx=2sec⁡2θ dθ\mathrm{d}x=2\sec^2\theta\,\mathrm{d}\theta. Missing that square changes the integral.

Use the plus-sign pattern at a new scale.

Back-substitution and transformed limits

Substitution can remove a root without making the answer immediate. With x=3sin⁡θx=3\sin\theta, the integral of 9−x2\sqrt{9-x^2} becomes the integral of 9cos⁡2θ9\cos^2\theta. The double-angle identity cos⁡2θ=(1+cos⁡2θ)/2\cos^2\theta=(1+\cos2\theta)/2 then makes it integrable.

For an indefinite integral, return every term to xx. For a definite integral, changing the limits allows you to finish entirely in θ\theta. These are two consistent ways to finish the same method.

Worked example 3: return an indefinite answer to x

Question. Find ∫9−x2 dx\displaystyle\int\sqrt{9-x^2}\,\mathrm{d}x, for −3<x<3-3<x<3.

Plan. Substitute x=3sin⁡θx=3\sin\theta, integrate using a double angle, then express both θ\theta and sin⁡2θ\sin2\theta in terms of xx.

∫9−x2 dx=9∫cos⁡2θ dθ9−x2=3cos⁡θ=92∫(1+cos⁡2θ) dθdouble-angle identity=92θ+94sin⁡2θ+C\begin{aligned} \int\sqrt{9-x^2}\,\mathrm{d}x &=9\int\cos^2\theta\,\mathrm{d}\theta && \sqrt{9-x^2}=3\cos\theta\\ &=\frac92\int(1+\cos2\theta)\,\mathrm{d}\theta && \text{double-angle identity}\\ &=\frac92\theta+\frac94\sin2\theta+C \end{aligned}

Use the chosen branch −π/2<θ<π/2-\pi/2<\theta<\pi/2. Then θ=sin⁡−1(x/3)\theta=\sin^{-1}(x/3), sin⁡θ=x/3\sin\theta=x/3, and cos⁡θ=9−x2/3\cos\theta=\sqrt{9-x^2}/3. Therefore

sin⁡2θ=2sin⁡θcos⁡θ=2x9−x29\sin2\theta=2\sin\theta\cos\theta =\frac{2x\sqrt{9-x^2}}9

Answer. x29−x2+92sin⁡−1(x/3)+C\frac{x}{2}\sqrt{9-x^2}+\frac92\sin^{-1}(x/3)+C.

Check. Differentiating gives

129−x2−x229−x2+929−x2=9−x2\frac12\sqrt{9-x^2}-\frac{x^2}{2\sqrt{9-x^2}} +\frac9{2\sqrt{9-x^2}} =\sqrt{9-x^2}

Keep the integrand unchanged, but now ask for a definite integral. The main new decision is what happens to the endpoints.

Worked example 4: finish with angle limits

Question. Find the exact value of ∫03/29−x2 dx\displaystyle\int_0^{3/2}\sqrt{9-x^2}\,\mathrm{d}x.

Plan. Use the same sine substitution and change both limits, so no back-substitution is needed.

x=0⇒θ=0,x=32⇒θ=π6∫03/29−x2 dx=9∫0π/6cos⁡2θ dθ=[92θ+94sin⁡2θ]0π/6integrate as before=3π4+938sin⁡(π/3)=3/2\begin{aligned} x=0&\Rightarrow\theta=0,\qquad x=\frac32\Rightarrow\theta=\frac{\pi}{6}\\ \int_0^{3/2}\sqrt{9-x^2}\,\mathrm{d}x &=9\int_0^{\pi/6}\cos^2\theta\,\mathrm{d}\theta\\ &=\left[\frac92\theta+\frac94\sin2\theta\right]_0^{\pi/6} && \text{integrate as before}\\ &=\frac{3\pi}{4}+\frac{9\sqrt3}{8} && \sin(\pi/3)=\sqrt3/2 \end{aligned}

Answer. 3π/4+93/83\pi/4+9\sqrt3/8.

Check. The curve decreases from 3 to 33/23\sqrt3/2 over a width of 3/23/2. Thus the area lies between 93/49\sqrt3/4 and 9/29/2. The answer is approximately 4.3054.305, within these bounds.

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