Unit 5, lesson 57

Trigonometric Integrals

How to integrate products of powers of sine and cosine, or of tangent and secant, by choosing a substitution from the parity of the exponents, and how product-to-sum identities handle different frequencies.

  • Theory part 113
  • Practice part 114
  • Integration Techniques

Theory, part 113

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Practice, part 114

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Integrals like ∫sin⁡3xcos⁡2x dx\int \sin^3 x \cos^2 x \, dx show up whenever waves are squared or multiplied, for example in the energy of a vibrating string or the power in an electric circuit. None of them needs a new method. A trigonometric identity rewrites the integrand so that a plain substitution finishes it, and the exponents tell you which identity to use.

Key terms

Trigonometric identities for integration

Definition For all real xx,

sin⁡2x+cos⁡2x=1,sin⁡2x=1−cos⁡2x2,cos⁡2x=1+cos⁡2x2,\sin^2 x + \cos^2 x = 1, \qquad \sin^2 x = \frac{1 - \cos 2x}{2}, \qquad \cos^2 x = \frac{1 + \cos 2x}{2},

and sin⁡xcos⁡x=12sin⁡2x\sin x \cos x = \frac{1}{2} \sin 2x. Wherever cos⁡x≠0\cos x \neq 0, also 1+tan⁡2x=sec⁡2x1 + \tan^2 x = \sec^2 x. The second and third formulas are the half-angle or power-reduction identities.

In plain words The Pythagorean identities trade one squared function for the other. The half-angle identities trade a square for a first power of a cosine with double the frequency, which is easy to integrate.

Example By the half-angle identity,

∫cos⁡2x dx=∫1+cos⁡2x2 dx=x2+sin⁡2x4+C.\int \cos^2 x \, dx = \int \frac{1 + \cos 2x}{2} \, dx = \frac{x}{2} + \frac{\sin 2x}{4} + C.

In the same way

∫0πsin⁡2x dx=[x2−sin⁡2x4]0π=π2.\int_0^{\pi} \sin^2 x \, dx = \Big[ \frac{x}{2} - \frac{\sin 2x}{4} \Big]_0^{\pi} = \frac{\pi}{2}.

Powers of sine and cosine

Definition Let mm and nn be nonnegative integers and consider ∫sin⁡mxcos⁡nx dx\int \sin^m x \cos^n x \, dx.

  1. If nn is odd, n=2k+1n = 2k + 1, write cos⁡nx=(1−sin⁡2x)kcos⁡x\cos^n x = (1 - \sin^2 x)^k \cos x and substitute u=sin⁡xu = \sin x, du=cos⁡x dxdu = \cos x \, dx. The integral becomes ∫um(1−u2)k du\int u^m (1 - u^2)^k \, du.
  2. If mm is odd, m=2k+1m = 2k + 1, write sin⁡mx=(1−cos⁡2x)ksin⁡x\sin^m x = (1 - \cos^2 x)^k \sin x and substitute u=cos⁡xu = \cos x, du=−sin⁡x dxdu = -\sin x \, dx. The integral becomes −∫(1−u2)kun du-\int (1 - u^2)^k u^n \, du.
  3. If mm and nn are both even, use the half-angle identities to lower the powers, repeating as needed.

In plain words An odd power lets you set aside one factor to be the dudu and turn everything else into the other function. With only even powers there is nothing to set aside, so you halve the powers instead.

Example In ∫sin⁡3xcos⁡2x dx\int \sin^3 x \cos^2 x \, dx the sine power is odd. Save one sin⁡x\sin x and let u=cos⁡xu = \cos x.

∫(1−cos⁡2x)cos⁡2xsin⁡x dx=−∫(1−u2)u2 du=−cos⁡3x3+cos⁡5x5+C.\int (1 - \cos^2 x) \cos^2 x \sin x \, dx = -\int (1 - u^2) u^2 \, du = -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C.

Powers of tangent and secant

Definition On any interval where cos⁡x≠0\cos x \neq 0,

∫tan⁡x dx=ln⁡∣sec⁡x∣+C,∫sec⁡x dx=ln⁡∣sec⁡x+tan⁡x∣+C.\int \tan x \, dx = \ln \lvert \sec x \rvert + C, \qquad \int \sec x \, dx = \ln \lvert \sec x + \tan x \rvert + C.

For ∫tan⁡mxsec⁡nx dx\int \tan^m x \sec^n x \, dx with nonnegative integers mm and nn,

  1. if nn is even and n≥2n \ge 2, save sec⁡2x\sec^2 x, write the remaining secants as powers of 1+tan⁡2x1 + \tan^2 x, and substitute u=tan⁡xu = \tan x, du=sec⁡2x dxdu = \sec^2 x \, dx,
  2. if mm is odd and n≥1n \ge 1, save sec⁡xtan⁡x\sec x \tan x, write the remaining tangents as powers of sec⁡2x−1\sec^2 x - 1, and substitute u=sec⁡xu = \sec x, du=sec⁡xtan⁡x dxdu = \sec x \tan x \, dx.

Other cases are handled with tan⁡2x=sec⁡2x−1\tan^2 x = \sec^2 x - 1, integration by parts, or a reduction formula.

In plain words The derivative of tan⁡x\tan x is sec⁡2x\sec^2 x and the derivative of sec⁡x\sec x is sec⁡xtan⁡x\sec x \tan x. Look for one of those two pieces in the integrand and make it your dudu.

Example In ∫tan⁡2xsec⁡4x dx\int \tan^2 x \sec^4 x \, dx the secant power is even. Save sec⁡2x\sec^2 x and let u=tan⁡xu = \tan x.

∫tan⁡2x(1+tan⁡2x)sec⁡2x dx=∫(u2+u4) du=tan⁡3x3+tan⁡5x5+C.\int \tan^2 x (1 + \tan^2 x) \sec^2 x \, dx = \int (u^2 + u^4) \, du = \frac{\tan^3 x}{3} + \frac{\tan^5 x}{5} + C.

Parity strategies

Definition The parity of the exponents, odd or even, decides the substitution. Here is the summary for the two families.

IntegrandConditionSaveSubstitute
sin⁡mxcos⁡nx\sin^m x \cos^n xnn oddcos⁡x dx\cos x \, dxu=sin⁡xu = \sin x
sin⁡mxcos⁡nx\sin^m x \cos^n xmm oddsin⁡x dx\sin x \, dxu=cos⁡xu = \cos x
sin⁡mxcos⁡nx\sin^m x \cos^n xboth evennothinghalf-angle identities
tan⁡mxsec⁡nx\tan^m x \sec^n xnn even, n≥2n \ge 2sec⁡2x dx\sec^2 x \, dxu=tan⁡xu = \tan x
tan⁡mxsec⁡nx\tan^m x \sec^n xmm odd, n≥1n \ge 1sec⁡xtan⁡x dx\sec x \tan x \, dxu=sec⁡xu = \sec x

When both conditions of a family hold, either choice works, and the answers differ by a constant.

In plain words Look at the exponents first. They tell you which factor to keep apart and which identity turns the rest into the right form.

Example In ∫tan⁡3xsec⁡x dx\int \tan^3 x \sec x \, dx the tangent power is odd and the secant power is at least 11. Save sec⁡xtan⁡x\sec x \tan x and let u=sec⁡xu = \sec x.

∫(sec⁡2x−1)sec⁡xtan⁡x dx=∫(u2−1) du=sec⁡3x3−sec⁡x+C.\int (\sec^2 x - 1) \sec x \tan x \, dx = \int (u^2 - 1) \, du = \frac{\sec^3 x}{3} - \sec x + C.

Product-to-sum identities

Definition For all real AA and BB,

sin⁡Acos⁡B=12[sin⁡(A−B)+sin⁡(A+B)],\sin A \cos B = \tfrac{1}{2} \big[ \sin(A - B) + \sin(A + B) \big],sin⁡Asin⁡B=12[cos⁡(A−B)−cos⁡(A+B)],\sin A \sin B = \tfrac{1}{2} \big[ \cos(A - B) - \cos(A + B) \big],cos⁡Acos⁡B=12[cos⁡(A−B)+cos⁡(A+B)].\cos A \cos B = \tfrac{1}{2} \big[ \cos(A - B) + \cos(A + B) \big].

In plain words A product of two waves with different frequencies is a sum of two waves, one at the difference and one at the sum of the frequencies. Sums are easy to integrate.

Example With A=3xA = 3x and B=2xB = 2x,

∫sin⁡3xcos⁡2x dx=12∫(sin⁡x+sin⁡5x) dx=−cos⁡x2−cos⁡5x10+C.\int \sin 3x \cos 2x \, dx = \frac{1}{2} \int (\sin x + \sin 5x) \, dx = -\frac{\cos x}{2} - \frac{\cos 5x}{10} + C.

Check yourself

  1. Find ∫cos⁡3x dx\int \cos^3 x \, dx.
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The cosine power is odd. Write cos⁡3x=(1−sin⁡2x)cos⁡x\cos^3 x = (1 - \sin^2 x) \cos x and let u=sin⁡xu = \sin x. The result is

∫(1−u2) du=sin⁡x−sin⁡3x3+C.\int (1 - u^2) \, du = \sin x - \frac{\sin^3 x}{3} + C.
  1. Evaluate ∫0π/2sin⁡2xcos⁡2x dx\int_0^{\pi/2} \sin^2 x \cos^2 x \, dx.
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Both powers are even, so lower them with the identities,

sin⁡2xcos⁡2x=14sin⁡22x=1−cos⁡4x8.\sin^2 x \cos^2 x = \frac{1}{4} \sin^2 2x = \frac{1 - \cos 4x}{8}.

Integrating from 00 to π2\frac{\pi}{2} gives

[x8−sin⁡4x32]0π/2=π16.\Big[ \frac{x}{8} - \frac{\sin 4x}{32} \Big]_0^{\pi/2} = \frac{\pi}{16}.
  1. Show that ∫−ππsin⁡2xsin⁡3x dx=0\int_{-\pi}^{\pi} \sin 2x \sin 3x \, dx = 0.
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The product-to-sum identity with A=2xA = 2x and B=3xB = 3x, together with cos⁡(−x)=cos⁡x\cos(-x) = \cos x, gives

sin⁡2xsin⁡3x=12(cos⁡x−cos⁡5x).\sin 2x \sin 3x = \frac{1}{2}(\cos x - \cos 5x).

An antiderivative is 12(sin⁡x−sin⁡5x5)\frac{1}{2} \left( \sin x - \frac{\sin 5x}{5} \right). Both sin⁡x\sin x and sin⁡5x5\frac{\sin 5x}{5} are zero at ±π\pm \pi, so the integral is 00.

How this lesson connects

It also uses trig identities.