Difficulty: Intermediate | Prerequisites: Comfort with basic integration rules, the Pythagorean identity, and u-substitution from earlier chapters.
Trigonometric integrals sit at the heart of Integral Calculus and appear heavily on exams. The central idea is that you rarely integrate powers of sin and cos directly; instead, you rewrite the integrand using identities (Pythagorean, double-angle, or reciprocal) until a clean substitution appears. This material builds directly on the u-substitution techniques from earlier in the course and feeds into later topics like trigonometric substitution for rational expressions. If you can spot which identity to apply and which variable to substitute, you can handle nearly every problem in this chapter.
Use Pythagorean and double-angle identities to rewrite trig integrands so that one trig function becomes the derivative of the other, then substitute. For odd powers, peel off one factor and convert the rest. For even powers, apply the half-angle (double-angle) formulas to reduce the degree.
Pythagorean identity
sin²(x) + cos²(x) = 1, rearranged as sin²(x) = 1 - cos²(x) or cos²(x) = 1 - sin²(x). Think of it as: the tool that lets you swap one trig function for the other inside an integral.
Double-angle / half-angle identity
cos(2x) = 1 - 2 sin²(x) = 2 cos²(x) - 1, so sin²(x) = (1 - cos(2x))/2 and cos²(x) = (1 + cos(2x))/2. In simple terms, this converts an even power of sin or cos into something you can integrate directly.
Trigonometric substitution (u-sub with trig)
Setting u equal to one trig function (e.g. u = cos(x)) so that du equals the other trig function times dx (du = -sin(x) dx). This is the standard move once you have separated the integrand into sin(x) f(cos(x)) or cos(x) f(sin(x)).
sec and tan relationship
(d/dx) sec(x) = sec(x) tan(x) and (d/dx) tan(x) = sec²(x). These derivatives drive the substitution strategy for integrands involving sec and tan.
Odd vs even power strategy
The classification that determines your approach: odd powers let you peel off one factor and substitute; even powers require the half-angle reduction formulas.
When the integrand contains an odd power of sin or cos, peel off one factor to pair with dx, then convert everything remaining using sin²(x) = 1 - cos²(x) (or vice versa).
sin(x) f(cos(x)) form: set u = cos(x), du = -sin(x) dx
Example (Q1): ∫ (2 sin θ - sin³ θ) dθ. Factor as sin θ (1 + cos² θ). Set u = cos θ, get ∫ (1 + u²) du = 4/3.
cos(x) f(sin(x)) form: set u = sin(x), du = cos(x) dx
Example (Q2): ∫ 3 sin²(x) cos³(x) dx. Rewrite cos³(x) = cos²(x) cos(x) = (1 - sin²(x)) cos(x). Set u = sin(x), get ∫ 3(u² - u⁴) du = 2/5.
The key recognition: an odd power always leaves one lone sin or cos to absorb into du.
When the integrand has only even powers, apply the half-angle identities to reduce the degree.
sin²(x) = (1 - cos(2x))/2
cos²(x) = (1 + cos(2x))/2
Examples:
(Q5): ∫ (1 + 4 sin²(θ)) dθ = ∫ (3 - 2 cos(2θ)) dθ = 3θ - sin(2θ). Evaluated from 0 to π/4 gives (3/4)π - 1.
(Q6): ∫ (5 cos²(x) + sin²(x)) dx. Rewrite as (1/2)(6 + 4 cos(2x)). Integrate to get (3/2)π.
Some integrands simplify by dividing out a power of cos and rewriting in terms of sec and tan.
Example (Q4): ∫ (3 cos x + 5 sin x) / cos³ x dx = ∫ (3 + 5 tan x) sec² x dx. Set u = tan x, du = sec² x dx. Result: 11/2.
When sec(x) tan(x) appears in the integrand, use u = sec(x) (or a variant like u = sec(4x)).
Example (Q8): ∫ sec⁶(4x) tan(4x) dx. Set u = sec(4x), du = 4 sec(4x) tan(4x) dx. Result: (1/24) sec⁶(4x) + C.
Example (Q9): ∫ sec(x) tan(x) / (5 + 2 sec(x)) dx. Set u = 5 + 2 sec(x), du = 2 sec(x) tan(x) dx. Result: (1/2) ln(9/7).
Sometimes the integrand mixes a derivative with trig terms. Split into manageable parts.
Example (Q10): ∫ sin(x)(f'(cos(x)) + 3 sin(x)) dx with f(0) = 1, f(1) = 5. Split into I₁ + I₂. For I₁, set u = cos(x) and apply the FTC to get f(1) - f(0) = 4. For I₂, use the half-angle identity to get (3/4)π. Total: 4 + (3/4)π.
\sin^2(x) = 1 - \cos^2(x)\cos^2(x) = 1 - \sin^2(x)\sin^2(x) = \frac{1}{2}\bigl(1 - \cos(2x)\bigr)\cos^2(x) = \frac{1}{2}\bigl(1 + \cos(2x)\bigr)\frac{d}{dx}\sec(x) = \sec(x)\tan(x)\frac{d}{dx}\tan(x) = \sec^2(x)Trigonometric integrals appear in electrical engineering when computing the RMS (root mean square) value of AC signals, which requires integrating sin²(t) or cos²(t) over a period. They also arise in physics when calculating work done by oscillating forces, or in computing areas and arc lengths of curves described by trig functions.
Students often forget to change the limits of integration when substituting u = cos(x) or u = sin(x). The original limits are in x; the new limits must be in u.
Applying the half-angle identity when the power is odd wastes time. Odd powers have a simpler route: peel off one factor and use the Pythagorean identity.
When peeling off sin(x) and setting u = cos(x), the sign flips: du = -sin(x) dx, which means the integral picks up a negative sign. Dropping that minus sign is one of the most common errors.
Students sometimes confuse (d/dx) sec(x) = sec(x) tan(x) with (d/dx) tan(x) = sec²(x). Mix them up and the entire substitution falls apart.
⚠️ Expect at least one problem requiring you to choose between the odd-power strategy and the half-angle strategy. The first step is always: check whether the total power of sin or cos is odd or even.
⚠️ Problems combining sec and tan (like Q4, Q8, Q9) appear frequently. Know both derivatives cold: (d/dx) sec(x) and (d/dx) tan(x).
⚠️ Composite problems (Q10) that mix the FTC with a trig substitution are a favourite exam question. They test whether you can split an integral into parts and handle each with the right technique.
⚠️ Always simplify your final answer. Exam graders may mark down an unsimplified expression even if the integration was correct.
True or False: To integrate sin³(x) cos²(x), you should use the half-angle identity. (False: the sin power is odd, so peel off sin(x) and convert.)
Fill in the blank: sin²(x) = ______ in terms of cos(2x). (1/2 (1 - cos(2x)))
True or False: If u = cos(x), then du = sin(x) dx. (False: du = -sin(x) dx.)
Fill in the blank: (d/dx) sec(x) = ______. (sec(x) tan(x))
True or False: When using u-substitution on a definite integral, you must convert the limits to the new variable. (True.)
Q: Evaluate ∫₀^{π/2} (2 sin θ - sin³ θ) dθ.
A: Factor as sin θ (1 + cos² θ), set u = cos θ, du = -sin θ dθ. Integral becomes ∫₀¹ (1 + u²) du = [u + u³/3]₀¹ = 4/3.
Q: Evaluate ∫₀^{π/2} 3 sin²(x) cos³(x) dx.
A: Rewrite cos³(x) = (1 - sin²(x)) cos(x). Set u = sin(x). Integral becomes ∫₀¹ 3(u² - u⁴) du = [u³ - (3/5)u⁵]₀¹ = 2/5.
Q: Evaluate ∫₀^{π/4} (1 + 4 sin²(θ)) dθ.
A: Use sin²(θ) = (1 - cos(2θ))/2. Integrand becomes 3 - 2 cos(2θ). Result: [3θ - sin(2θ)]₀^{π/4} = (3/4)π - 1.
Q: Evaluate ∫₀^{π/4} (3 cos x + 5 sin x) / cos³(x) dx.
A: Divide to get (3 + 5 tan x) sec²(x). Set u = tan x. ∫₀¹ (3 + 5u) du = [3u + (5/2)u²]₀¹ = 11/2.
Q: Determine ∫ sec⁶(4x) tan(4x) dx.
A: Set u = sec(4x), du = 4 sec(4x) tan(4x) dx. Integral = (1/4) ∫ u⁵ du = (1/24) sec⁶(4x) + C.
This material connects directly to trigonometric substitution for rational expressions (coming up next), where you reverse the process: instead of substituting away from trig, you substitute into trig. It also feeds into integration by parts, since some mixed trig integrals require combining both techniques. The volume-of-revolution problems later in this homework use these same trig integrals once the disk/washer formula is set up.
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