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Trigonometric Integrals
Calculus 2 · Chapter 1: Review & Advanced Integration TechniquesSection 1.3 · lesson 3 of 5411 min · free
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In this lesson
Read the parity, pick the strategy, and the integral falls apart.
- Use parity to choose between peeling off a factor and half-angle reduction
- Handle \tan^{m}x\sec^{n}x through the even-secant and odd-tangent cases
- Convert mismatched arguments with the product-to-sum identities
- Step high even powers down with a reduction formula
Transcript
Everything said in the video, section by section (1,653 words). Read along, or search it.
MAT-265 - Calculus 2 · Section 1.3
Trigonometric Integrals
An audio engineer is mixing two pure tones. A four-forty hertz concert A, and a slightly detuned four-forty-two hertz note. They need to know the average power delivered to the speaker over one cycle. That calculation comes down to integrating a product of sines and cosines. Sometimes those factors share an argument. Sometimes they don't. And that difference turns out to matter enormously. A physicist finding the root-mean-square voltage of household AC power runs into the same kind of integral. None of these yield to a single substitution. So this section builds you a decision-making toolkit. Read the powers, pick the matching strategy, and the integral falls apart cleanly.
The Toolkit
Parity Decides Everything
Trigonometric integrals are antiderivatives whose integrands are products of trig functions raised to integer powers. And here's the thing worth understanding up front. The challenge is almost never the integration itself. It's recognizing which algebraic identity turns the integrand into something a simple u-substitution can finish. Every strategy in this section is driven by one single feature. Parity. Whether your exponents are odd or even. Get in the habit of looking at the exponents before you do anything else.
Strategy 1
An Odd Power: Peel, Convert, Substitute
Start with powers of sine and cosine, and suppose at least one exponent is odd. Here's the move. Peel off a single factor of that function to pair with d x. Convert what's left using the Pythagorean identity. Then substitute. The rule is worth stating precisely. If the power of cosine is odd, save one cosine and let u be sine x. If the power of sine is odd, save one sine and let u be cosine x. Either way, that saved factor becomes part of d u. That's why you save it. Let's see it. The integral of sine to the fourth, cosine cubed. Cosine has power three — odd. So save one cosine, and rewrite the remaining cosine squared as one minus sine squared. Now let u be sine x. The saved cosine d x is exactly d u. And what's left is a polynomial. u to the fourth minus u to the sixth. That integrates on sight. u to the fifth over five, minus u to the seventh over seven. Back-substitute and you're done. One note. If both powers are odd, either choice works. Pick the one whose saved factor leaves you the simpler polynomial.
Strategy 2
Both Powers Even: Half-Angle Reduction
Now what happens when both exponents are even? You're stuck. There's no single factor you can peel off to form a clean d u. So you reduce the powers instead, using the half-angle identities. Sine squared x is one minus cosine two x, all over two. Cosine squared x is one plus cosine two x, over two. Each application halves the exponent — at the cost of doubling the angle. That's the trade. Let's do the one every engineer meets. The integral of sine squared. Household AC voltage is modeled as V naught sine omega t, and average power depends on the mean of V squared. Apply the identity and you get one half the integral of one, minus one half the integral of cosine two x. The answer is x over two, minus sine two x over four, plus C. Now look at what that means physically. Over one full period, the sine two x term averages to zero. Which leaves an average value of one half. And that is exactly why RMS voltage is V naught over root two — about a hundred and twenty volts for a hundred and seventy volt peak. For higher even powers you just apply the identity again. Cosine to the fourth becomes cosine squared, squared — reduce, then reduce the leftover cosine squared two x a second time.
Tangent and Secant
Two Clean Cases Cover Most Problems
Now switch families. Powers of tangent and secant. These rely on the companion Pythagorean identity: one plus tangent squared equals secant squared. Plus two derivative facts. The derivative of tangent is secant squared. The derivative of secant is secant tangent. Two clean cases cover most problems, and they mirror each other exactly. First case. If the power of secant is even, and at least two, save a factor of secant squared to serve as d u. Convert the remaining even secant power to tangents, and let u be tangent x. Watch it on the integral of tangent squared secant to the fourth. Save one secant squared, convert the other into one plus tangent squared. You get the integral of u squared times one plus u squared. Elementary. Second case. If the power of tangent is odd, and at least one secant factor is present, save a factor of secant tangent instead — because that's the derivative of secant. Convert the leftover even tangent power using tangent squared equals secant squared minus one, and let u be secant x. Two awkward cases fall outside these patterns, and they're worth memorizing as anchors. Odd secant with even tangent — like the integral of secant cubed — usually needs integration by parts. And the integral of tangent x is the natural log of absolute secant x, plus C.
Reference
The Strategy Selector
Let's put all of that in one place, because in practice this is how you'll actually use it. Look at the integrand. Read the powers. Then read off the method. Sine and cosine, with cosine odd: save a cosine, let u be sine. Sine and cosine, with sine odd: save a sine, let u be cosine. Sine and cosine, both even: power reduction, using the half-angle identities. Tangent and secant, secant even: save secant squared, let u be tangent. Tangent and secant, tangent odd: save secant tangent, let u be secant. Five rows. That's the whole section. Here's something to sit with. In Example One the sine power was even and the cosine power odd, so the method keyed on the odd cosine. But suppose both had been odd — the integral of sine cubed cosine cubed. Why does either substitution work there? Try both, and compare the polynomials in u you end up having to integrate. One of them is friendlier.
Different Arguments
When the Angles Do Not Match
Everything so far assumed your factors share the same angle. So what happens when they don't? Back to the audio engineer, with sine of four-forty pi t against cosine of four-forty-two pi t. Here the parity tricks are useless. Not harder — useless. Because there's no shared Pythagorean structure to exploit. Sine squared plus cosine squared equals one needs both to be the same angle. So you do something different. You convert the product into a sum. Sine A cosine B is one half, sine of A plus B, plus sine of A minus B. There are companions for sine times sine, and cosine times cosine. Each one turns a stubborn product into a sum of single-argument sines or cosines — every one of which integrates immediately. Try the integral of sine five x cosine three x. Use the first identity with A equals five x and B equals three x. The product becomes one half, sine eight x plus sine two x. Integrate term by term and you're finished. And this identity is also why beats exist in acoustics. Two close frequencies combine into a fast oscillation modulated by a slow envelope — precisely the A plus B and A minus B terms the identity separates out.
Strategy 4
Reduction Formulas for High Powers
One more tool, for when the exponents get large. Repeated half-angle expansion works, but it becomes genuinely tedious by the time you're at the sixth or eighth power. A reduction formula fixes that. It's derived once, using integration by parts. And it expresses an integral with exponent n in terms of the same integral with exponent n minus two. So each application steps you down by two, and you apply it until what's left is elementary. For sine to the fourth: with n equals four, you get minus sine cubed x cosine x over four, plus three quarters of the integral of sine squared. And you already know the integral of sine squared from a moment ago. Substitute it in and simplify. That's it. Two lines, instead of expanding half-angles twice.
Avoiding Errors
Almost Every Mistake Is a Trigger You Did Not Check
Almost every mistake in this section traces to one thing. Applying a method whose trigger condition wasn't actually met. So let's name the four traps. One. Forcing the odd-power method when both powers are even. Try to peel a factor off the integral of sine squared cosine squared, and the leftover power is odd — the Pythagorean identity can't clean that up. You'll get stuck. Both even always means half-angle. Two. Reaching for half-angle when an odd power is sitting right there. It still works, but only after extra effort, whereas the odd-power method finishes in two lines. Always check parity first. Three. Confusing the secant and tangent triggers. The even-secant method needs an even secant power, so that a secant squared can be set aside. The odd-tangent method needs an odd tangent power, so that secant tangent can be. Mix them up — save a secant squared when secant is odd — and you're left with a half-power of secant that no identity removes. Four. Applying parity tricks across different arguments. That never works. Only product-to-sum does.
Key Takeaways
What to Carry Forward
So, five things to carry out of this section. Read the parity first. For sine to the m, cosine to the n, an odd exponent signals peel-and-substitute. Both even signals half-angle reduction. The saved factor must match a derivative. Cosine for u equals sine. Sine for u equals cosine. Secant squared for u equals tangent. Secant tangent for u equals secant. Even secant or odd tangent are the two clean tangent-secant cases. Odd secant with even tangent usually needs integration by parts. Different arguments demand product-to-sum identities. Parity strategies simply do not apply across mismatched angles. And reduction formulas tame high even powers efficiently, stepping the exponent down by two each time. Next up is section one point four, Trigonometric Substitution — where instead of simplifying a trig function, you deliberately introduce one to kill a radical.
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