Source: University of Texas at Austin, Calc 2 lecture notes
Tags: trig identities, trigonometric integrals, powers of sine, powers of cosine, half-angle formulas, Pythagorean identity, u-substitution, Calc 2
Difficulty: Intermediate Prerequisites: Basic trigonometric functions, Calc 1 integration techniques (u-substitution), familiarity with the unit circle.
Trigonometric identities are the backbone of evaluating integrals involving powers and products of sine, cosine, secant and tangent. In Calc 2, you move beyond simple antiderivatives into integrals that require algebraic manipulation before you can integrate. This topic sits right after basic integration techniques (u-substitution, integration by parts) and before trigonometric substitution and partial fractions. If you are comfortable rewriting trig expressions using identities and performing u-substitution, you are ready for this material.
Four core trig identities let you reduce complicated trig integrals into forms you can handle with u-substitution. Odd powers of sine or cosine: peel off one factor, convert the rest using sin²x + cos²x = 1, then substitute. Even powers: use the half-angle (power-reduction) identities to lower the degree until you can integrate directly.
Pythagorean identity
sin²x + cos²x = 1. The foundational relationship between sine and cosine on the unit circle. In simple terms, this means knowing one of sin x or cos x lets you find the other.
Power-reduction identity (half-angle identity)
Formulas that replace sin²x or cos²x with expressions involving cos(2x), cutting the power in half. Think of it as trading a squared trig function for a simpler cosine of double the angle.
u-substitution
A technique where you set u equal to part of the integrand so the remaining piece becomes du, simplifying the integral. In simple terms, you pick one chunk to rename and rewrite everything else in terms of that chunk.
Odd power strategy
When the power of sine or cosine is odd, factor out one copy of that function, convert the rest using sin²x + cos²x = 1, then substitute. Think of it as peeling off the factor you need for du, then rewriting what remains.
Even power strategy
When the power of sine or cosine is even, apply the half-angle identities repeatedly until every term can be integrated directly. In simple terms, you keep halving the exponents until there are no squared trig functions left.
Every trig integral in this unit comes back to four identities. Learn these cold.
Identity 1: Pythagorean identity
sin²x + cos²x = 1
Rearranged forms you will use constantly:
sin²x = 1 – cos²x (when you want to eliminate sine)
cos²x = 1 – sin²x (when you want to eliminate cosine)
Identity 2: Sine half-angle (power-reduction)
sin²x = (1 – cos(2x)) / 2
Used whenever you have an even power of sine and no odd factor to peel off.
Identity 3: Cosine half-angle (power-reduction)
cos²x = (1 + cos(2x)) / 2
The companion to Identity 2. Same purpose, applied to even powers of cosine.
Identity 4: Secant-tangent Pythagorean identity
sec²x = 1 + tan²x
This one drives the entire secant-tangent integration strategy (covered in Part 2 of these notes). It lets you convert between sec²x and tan²x freely.
The strategy here relies on having an odd exponent, which means you can always factor out one copy of the function to serve as du.
Method (odd power of cosine)
Factor out one cos x to pair with dx (this becomes du when u = sin x).
Rewrite the remaining cos²x factors as (1 – sin²x) using Identity 1.
Set u = sin x, du = cos x dx.
You now have a polynomial in u. Integrate term by term.
Worked example: ∫ cos³x dx
Split: ∫ cos²x · cos x dx
Rewrite: ∫ (1 – sin²x) · cos x dx
Substitute u = sin x, du = cos x dx:
∫ (1 – u²) du = u – u³/3 + C = sin x – sin³x / 3 + C
The same method works for odd powers of sine: factor out one sin x, convert the rest to cosine using sin²x = 1 – cos²x, and set u = cos x (note du = –sin x dx, so watch the sign).
With even powers you cannot peel off a single factor for u-substitution, because that would leave an odd power of the other function with no clean substitution. Instead, use the half-angle identities to reduce the powers.
Method (even power of sine)
Replace sin²x with (1 – cos(2x)) / 2 everywhere.
Expand and simplify.
If any squared trig terms remain (e.g. cos²(2x)), apply the half-angle identity again.
Integrate term by term.
Worked example: ∫ sin⁴x dx
Rewrite sin⁴x as (sin²x)²:
(sin²x)² = ((1 – cos(2x)) / 2)² = (1 – 2cos(2x) + cos²(2x)) / 4
The cos²(2x) term still has an even power, so apply Identity 3 again:
cos²(2x) = (1 + cos(4x)) / 2
Substitute back:
(1/4)(1 – 2cos(2x) + (1 + cos(4x))/2) = (1/4)(3/2 – 2cos(2x) + cos(4x)/2)
Integrate each term:
(1/4)(3x/2 – sin(2x) + sin(4x)/8) + C
The same strategy applies to even powers of cosine: use cos²x = (1 + cos(2x)) / 2 and repeat until no squared terms remain.
Identity | Formula | When to use |
|---|---|---|
Pythagorean | sin²x + cos²x = 1 | Odd powers: convert leftover factors |
Sine half-angle | sin²x = (1 – cos 2x) / 2 | Even powers of sine |
Cosine half-angle | cos²x = (1 + cos 2x) / 2 | Even powers of cosine |
Sec-tan Pythagorean | sec²x = 1 + tan²x | Secant/tangent integrals (Part 2) |
Trigonometric integrals appear whenever periodic signals need to be decomposed or averaged. In electrical engineering, computing the RMS (root mean square) value of an AC voltage requires integrating sin²(ωt) over a period, which uses exactly the half-angle identity from this unit. In physics, calculating the energy stored in oscillating systems (springs, pendulums, electromagnetic waves) involves the same power-reduction techniques.
Students often try to use half-angle identities on odd powers. You do not need them: the odd-power method (peel and substitute) is faster and cleaner.
Students frequently forget the factor of 1/2 in the half-angle identities. sin²x = (1 – cos 2x) / 2, not (1 – cos 2x).
When substituting u = sin x in an odd-cosine integral, some students forget that du = cos x dx and try to substitute without the companion factor.
Students sometimes confuse cos(2x) with 2cos(x). These are different functions. cos(2x) = 2cos²x – 1, not 2cos x.
⚠️ Professors commonly test whether you can choose the right strategy (odd vs even power). Read the exponent before you start writing.
⚠️ A favourite exam question is ∫ sin²x dx or ∫ cos²x dx on its own, testing whether you know the half-angle identity by heart.
⚠️ Expect at least one problem combining both strategies, such as ∫ sin³x cos²x dx, where you peel off one sin x and convert the remaining sin²x.
⚠️ Double-check your final answer by differentiating it. If you get back the original integrand, you are correct.
True or false: To integrate cos⁵x, you should use the half-angle identity. (False. Odd power: peel and substitute.)
Fill in the blank: sin²x = ______ . ((1 – cos 2x) / 2)
True or false: The substitution u = sin x gives du = sin x dx. (False. du = cos x dx.)
Fill in the blank: When integrating sin⁴x, rewrite it as (______)². ((sin²x)²)
True or false: cos²(2x) can be reduced further using the half-angle identity. (True. cos²(2x) = (1 + cos 4x) / 2.)
Q: Evaluate ∫ cos³x dx.
A: Factor out cos x, rewrite cos²x = 1 – sin²x, substitute u = sin x. Result: sin x – sin³x / 3 + C.
Q: Evaluate ∫ sin²x dx.
A: Apply the half-angle identity: ∫ (1 – cos 2x)/2 dx = x/2 – sin(2x)/4 + C.
Q: What is the first step when evaluating ∫ sin⁵x cos²x dx?
A: The sine has an odd power. Factor out one sin x, convert sin⁴x = (1 – cos²x)², then set u = cos x, du = –sin x dx.
Q: Evaluate ∫ sin⁴x dx.
A: Apply the half-angle identity twice. Final answer: (1/4)(3x/2 – sin(2x) + sin(4x)/8) + C.
Q: A student writes ∫ cos²x dx = sin x cos x + C. What is wrong?
A: They tried to guess the antiderivative instead of using the half-angle identity. The correct result is x/2 + sin(2x)/4 + C. Differentiating the student's answer gives cos²x – sin²x, which is cos(2x), not cos²x.
This material connects directly to trigonometric substitution (the next major Calc 2 topic), where you substitute x = sin θ or x = tan θ and then integrate the resulting trig expressions using the techniques covered here. It also feeds into Fourier series and Fourier transforms, where integrals of sin²(nx) and cos²(nx) over a period are computed constantly. If you go on to differential equations, the same identities appear when solving second-order equations with oscillatory solutions.
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