Source: Lecture 3, Section 7.2
Tags: trigonometric integrals, sin cos powers, trig identities, half-angle formulas, u-substitution, reduction, Calculus II, NYU
Difficulty: Intermediate | Prerequisites: Basic integration techniques, u-substitution, Pythagorean identities from precalculus
Trigonometric integrals form a major topic in Calculus II and appear in nearly every exam. The techniques here build on u-substitution, which you should already be comfortable with, and rely heavily on a small set of trig identities. The core skill is recognising which identity to apply based on the powers of sine and cosine in the integrand. This section also lays the groundwork for trigonometric substitution (Section 7.3), where these integral results get used as building blocks.
When integrating powers of sin and cos, odd powers let you peel off one factor and use u-substitution via the Pythagorean identity. Even powers require half-angle identities to reduce the power. Products of sin(ax) and cos(bx) with different arguments use product-to-sum formulas. Tan and sec integrals follow their own parallel set of rules.
Pythagorean Identity
sin squared(x) + cos squared(x) = 1. The single most important identity in this section. Every odd-power trig integral uses it.
Half-angle Identities (Power-Reduction Formulas)
sin squared(x) = (1 minus cos(2x))/2 and cos squared(x) = (1 + cos(2x))/2. These convert even powers of sin or cos into expressions involving cos(2x), which you can then integrate directly. Think of them as the tool for even-power problems.
Double-Angle Identity
sin(2x) = 2 sin(x) cos(x). Useful when simplifying products of sin and cos with the same argument.
Product-to-Sum Formulas
Identities that convert products like sin(ax) cos(bx) into sums of sine or cosine terms with different arguments. These are used when the two trig functions have different frequencies (different multiples of x).
Reduction Formula
A recursive formula that expresses an integral of a trig function raised to the nth power in terms of the same integral with a lower power. Useful as a shortcut once derived, but you need to be able to derive them from scratch using the methods in this section.
You will use these identities constantly. They should be committed to memory.
Pythagorean family
sin squared(x) + cos squared(x) = 1
1 + tan squared(x) = sec squared(x)
1 + cot squared(x) = csc squared(x)
Reciprocal and quotient identities
tan(x) = sin(x)/cos(x), cot(x) = cos(x)/sin(x)
sec(x) = 1/cos(x), csc(x) = 1/sin(x)
cot(x) times tan(x) = 1
Double-angle and half-angle
sin(2x) = 2 sin(x) cos(x)
cos(2x) = cos squared(x) minus sin squared(x) = 1 minus 2 sin squared(x) = 2 cos squared(x) minus 1
sin squared(x) = (1 minus cos(2x))/2
cos squared(x) = (1 + cos(2x))/2
The last two (the half-angle/power-reduction forms) are the workhorses for even-power integrals. Exams will not give you these; you need them from memory.
This covers integrals of the form: integral of sin^n(x) dx or integral of cos^n(x) dx, on their own.
Odd power (e.g. sin cubed(x))
Strategy: peel off one factor of sin(x) or cos(x) to pair with dx, then convert the remaining even power using sin squared(x) = 1 minus cos squared(x).
Worked example: integral of sin cubed(x) dx.
Rewrite as integral of sin squared(x) times sin(x) dx.
Let u = cos(x), du = negative sin(x) dx, so sin(x) dx = negative du.
Replace sin squared(x) with 1 minus u squared.
Integral becomes negative integral of (1 minus u squared) du = negative u + u cubed / 3 + C.
Back-substitute: negative cos(x) + cos cubed(x)/3 + C.
Even power (e.g. cos to the fourth(x))
Strategy: use the half-angle identity repeatedly to eliminate all powers.
Worked example: integral of cos^4(x) dx.
Write cos^4(x) = (cos squared(x)) squared = ((1 + cos(2x))/2) squared.
Expand: (1 + 2 cos(2x) + cos squared(2x))/4.
Apply the half-angle identity again to cos squared(2x) = (1 + cos(4x))/2.
Result: 3/8 + (1/2) cos(2x) + (1/8) cos(4x).
Integrate term by term: 3x/8 + sin(2x)/4 + sin(4x)/32 + C.
Equivalently: (12x + 8 sin(2x) + sin(4x))/32 + C.
The key distinction: odd power means u-substitution with a Pythagorean identity; even power means half-angle identities applied until no powers remain.
These integrals have both sin and cos raised to powers. The strategy depends on whether either power is odd.
At least one odd power
Peel off one factor of whichever trig function has the odd power, then convert the rest using the Pythagorean identity.
Worked example: integral of sin^5(x) cos squared(x) dx.
sin has the odd power (5), so peel off one sin(x): rewrite as sin^4(x) cos squared(x) sin(x) dx.
Let u = cos(x), du = negative sin(x) dx.
Replace sin^4(x) = (sin squared(x)) squared = (1 minus cos squared(x)) squared = (1 minus u squared) squared.
The integral becomes negative integral of (1 minus u squared) squared times u squared du.
Expand: negative integral of u squared minus 2u^4 + u^6 du.
Integrate: negative (u cubed/3 minus 2u^5/5 + u^7/7) + C.
Back-substitute: negative cos cubed(x)/3 + 2 cos^5(x)/5 minus cos^7(x)/7 + C.
Both even powers
Use half-angle identities to reduce both.
Worked example: integral of sin squared(x) cos^4(x) dx.
Rewrite sin squared(x) = 1 minus cos squared(x), giving integral of cos^4(x) minus cos^6(x) dx.
Then apply the half-angle approach to each term separately.
This can get lengthy. The lecture notes show cos^4(x) expanded as 1/4 + (1/2) cos(2x) + (1/4) cos squared(2x), and so on.
Both odd powers
You have a choice: peel off from either factor. Pick whichever makes the algebra simpler.
Worked example: integral of sin cubed(x) cos^5(x) dx.
Peel off sin(x), let u = cos(x).
Integral becomes integral of u cubed (1 minus u squared) squared du (after converting sin squared to 1 minus u squared and adjusting signs).
Expand and integrate: sin^4(x)/4 minus sin^6(x)/3 + sin^8(x)/8 + C (expressed in terms of the original variable after back-substitution).
These integrals have their own set of standard results and techniques.
Standard results to memorise
Integral of tan(x) dx = ln|sec(x)| + C.
Derivation: rewrite tan(x) as sin(x)/cos(x), set u = cos(x), du = negative sin(x) dx. The integral becomes negative ln|cos(x)| + C = ln|sec(x)| + C.
Integral of sec(x) dx = ln|sec(x) + tan(x)| + C.
Derivation: multiply top and bottom by (sec(x) + tan(x)), then set u = sec(x) + tan(x), du = (sec(x) tan(x) + sec squared(x)) dx.
Integral of tan^m(x) sec^n(x) dx
If sec has an even power (n is even, n >= 2): peel off sec squared(x), set u = tan(x), du = sec squared(x) dx, and convert remaining sec squared factors using sec squared = 1 + tan squared.
Example: integral of tan^6(x) sec^4(x) dx. Rewrite as tan^6(x) sec squared(x) sec squared(x) dx. Peel off sec squared(x) for du, convert the other sec squared to (1 + tan squared). Set u = tan(x). Integral becomes integral of u^6 (1 + u squared) du = tan^7(x)/7 + tan^9(x)/9 + C.
If tan has an odd power: peel off sec(x) tan(x), set u = sec(x), du = sec(x) tan(x) dx, and convert remaining tan squared factors using tan squared = sec squared minus 1.
Integral of tan cubed(x) dx
Rewrite as tan(x) times tan squared(x) = tan(x)(sec squared(x) minus 1).
Split: integral of tan(x) sec squared(x) dx minus integral of tan(x) dx.
First part: u = tan(x), du = sec squared(x) dx, giving tan squared(x)/2.
Second part: ln|sec(x)|.
Result: tan squared(x)/2 minus ln|sec(x)| + C.
Integral of sec cubed(x) dx
This requires integration by parts. Set u = sec(x), dv = sec squared(x) dx.
After the parts formula and some algebraic rearrangement (the integral of sec cubed reappears on the right), you solve for it:
Result: (1/2)(sec(x) tan(x) + ln|sec(x) + tan(x)|) + C.
This result appears often enough that some courses expect you to have it memorised.
When the integrand is a product of sin and cos with different arguments (different multiples of x), use the product-to-sum identities.
The three formulas
sin(ax) sin(bx) = (1/2)[cos(a minus b)x minus cos(a + b)x]
cos(ax) cos(bx) = (1/2)[cos(a + b)x + cos(a minus b)x]
sin(ax) cos(bx) = (1/2)[sin(a + b)x + sin(a minus b)x]
These convert a product into a sum, which is straightforward to integrate term by term.
Worked example: integral of sin(4x) cos(5x) dx
Apply the third formula: sin(4x) cos(5x) = (1/2)[sin(9x) + sin(negative x)] = (1/2)[sin(9x) minus sin(x)].
Integrate: (1/2)[negative cos(9x)/9 + cos(x)] + C.
Simplify: negative cos(9x)/18 + cos(x)/2 + C.
These problems are mechanical once you know the formulas. The only challenge is keeping the signs and coefficients straight.
Students often try to use u-substitution on even-power trig integrals. It does not work cleanly because you cannot peel off a single sin(x) or cos(x) and still have a clean even power left. Even powers require the half-angle identity, full stop.
Students sometimes apply the half-angle identity to odd-power problems, making them far more complicated than necessary. If either sin or cos has an odd power, use the Pythagorean substitution approach instead.
When integrating sec(x), students frequently forget the multiply-by-(sec + tan)/(sec + tan) trick and get stuck. This is a standard technique worth memorising as a pattern.
In product-to-sum problems, students often make sign errors when a minus b is negative. Remember: sin(negative theta) = negative sin(theta) and cos(negative theta) = cos(theta).
Trig integrals are among the most heavily tested topics in Calculus II. Expect at least one on every exam.
The odd-power versus even-power decision is the single most important judgement call in this section. If you get the strategy wrong, the problem does not simplify.
Integral of sec cubed(x) dx is a classic exam question, often appearing either on its own or as part of a larger problem (e.g. after a trig substitution).
Product-to-sum formula questions are quick marks if you know the formulas. They are often worth practising last, since the method is purely mechanical.
True or False: To integrate sin^4(x), you should peel off a sin(x) and use u = cos(x). (Answer: False. sin^4 is an even power; use half-angle identities.)
Fill in the blank: sin squared(x) = ________. (Answer: (1 minus cos(2x))/2.)
True or False: Integral of tan(x) dx = negative ln|cos(x)| + C. (Answer: True. This is equivalent to ln|sec(x)| + C.)
Fill in the blank: When integrating sin^m(x) cos^n(x) and m is odd, set u = ________. (Answer: cos(x).)
True or False: The product-to-sum formulas are only needed when sin and cos have the same argument. (Answer: False. They are specifically for different arguments, like sin(3x) cos(5x).)
Q: Evaluate the integral of sin^5(x) dx.
A: Peel off sin(x), set u = cos(x). Rewrite sin^4(x) = (1 minus u squared) squared. Expand, integrate, back-substitute. Result: negative cos(x) + (2/3) cos cubed(x) minus cos^5(x)/5 + C.
Q: Evaluate the integral of cos squared(x) sin squared(x) dx.
A: Both powers are even. Use sin(x) cos(x) = sin(2x)/2, so the integrand = sin squared(2x)/4. Apply the half-angle identity: (1 minus cos(4x))/8. Integrate: x/8 minus sin(4x)/32 + C.
Q: Evaluate the integral of tan^4(x) sec squared(x) dx.
A: Set u = tan(x), du = sec squared(x) dx. The integral becomes integral of u^4 du = u^5/5 + C = tan^5(x)/5 + C.
Q: Evaluate the integral of cos(3x) cos(7x) dx.
A: Use the product-to-sum formula: cos(3x) cos(7x) = (1/2)[cos(10x) + cos(4x)]. Integrate: sin(10x)/20 + sin(4x)/8 + C (with the 1/2 factor distributed).
Q: Evaluate the integral of sec cubed(x) dx.
A: Use integration by parts with u = sec(x), dv = sec squared(x) dx. After the parts formula, the integral of sec cubed reappears on the right side. Solve algebraically: (1/2)(sec(x) tan(x) + ln|sec(x) + tan(x)|) + C.
Trigonometric integrals feed directly into trigonometric substitution (Section 7.3), where you replace algebraic expressions like sqrt(a squared minus x squared) with trig functions and then integrate the resulting trig expression using these techniques. The integral of sec cubed(x), for instance, appears constantly after trig substitution. These methods also arise in Fourier analysis and signal processing, where decomposing functions into sums of sines and cosines is the central idea.
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