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Trig Integrals — Definition, Formula & Examples

Trig integrals are integrals whose integrands involve trigonometric functions such as sin, cos, tan, sec, csc, and cot. Evaluating them requires specific formulas, identities, and techniques like substitution and reduction.

A trigonometric integral is any integral of the form ∫f(sin⁡x,cos⁡x,tan⁡x,sec⁡x,csc⁡x,cot⁡x) dx\int f(\sin x, \cos x, \tan x, \sec x, \csc x, \cot x)\,dx, where the integrand is composed of one or more trigonometric functions. Standard results include ∫sin⁡x dx=−cos⁡x+C\int \sin x\,dx = -\cos x + C, ∫cos⁡x dx=sin⁡x+C\int \cos x\,dx = \sin x + C, ∫sec⁡2x dx=tan⁡x+C\int \sec^2 x\,dx = \tan x + C, and ∫csc⁡2x dx=−cot⁡x+C\int \csc^2 x\,dx = -\cot x + C, among others. More complex forms are handled through trigonometric identities, uu-substitution, or integration by parts.

Key Formula

∫sin⁡x dx=−cos⁡x+C∫cos⁡x dx=sin⁡x+C∫tan⁡x dx=−ln⁡∣cos⁡x∣+C∫cot⁡x dx=ln⁡∣sin⁡x∣+C∫sec⁡x dx=ln⁡∣sec⁡x+tan⁡x∣+C∫csc⁡x dx=−ln⁡∣csc⁡x+cot⁡x∣+C∫sec⁡2x dx=tan⁡x+C∫csc⁡2x dx=−cot⁡x+C∫sec⁡xtan⁡x dx=sec⁡x+C∫csc⁡xcot⁡x dx=−csc⁡x+C\begin{aligned} \int \sin x\,dx &= -\cos x + C \\ \int \cos x\,dx &= \sin x + C \\ \int \tan x\,dx &= -\ln|\cos x| + C \\ \int \cot x\,dx &= \ln|\sin x| + C \\ \int \sec x\,dx &= \ln|\sec x + \tan x| + C \\ \int \csc x\,dx &= -\ln|\csc x + \cot x| + C \\ \int \sec^2 x\,dx &= \tan x + C \\ \int \csc^2 x\,dx &= -\cot x + C \\ \int \sec x \tan x\,dx &= \sec x + C \\ \int \csc x \cot x\,dx &= -\csc x + C \end{aligned}
Where:
  • xx = The variable of integration (angle, typically in radians)
  • CC = The constant of integration

How It Works

Start by checking whether your integral matches one of the basic trig integral formulas. If it does not match directly, apply a trigonometric identity to rewrite the integrand into a more manageable form. For powers of sine and cosine, use the identities sin⁡2x=1−cos⁡2x2\sin^2 x = \frac{1 - \cos 2x}{2} and cos⁡2x=1+cos⁡2x2\cos^2 x = \frac{1 + \cos 2x}{2} (power-reduction), or peel off one factor and substitute u=cos⁡xu = \cos x or u=sin⁡xu = \sin x when one exponent is odd. For products like sin⁡mxcos⁡nx\sin^m x \cos^n x, the strategy depends on whether mm or nn is odd. For sec⁡\sec and tan⁡\tan combinations, similar peel-and-substitute strategies apply using u=sec⁡xu = \sec x or u=tan⁡xu = \tan x.

Worked Example

Problem: Evaluate ∫sin⁡3xcos⁡2x dx\int \sin^3 x \cos^2 x\,dx.
Step 1: Since the power of sin is odd, peel off one factor of sin x and convert the remaining sin²x using the identity sin²x = 1 − cos²x.
∫sin⁡3xcos⁡2x dx=∫(1−cos⁡2x)cos⁡2xsin⁡x dx\int \sin^3 x \cos^2 x\,dx = \int (1 - \cos^2 x)\cos^2 x \sin x\,dx
Step 2: Substitute u = cos x, so du = −sin x dx.
=−∫(1−u2)u2 du=−∫(u2−u4) du= -\int (1 - u^2)u^2\,du = -\int (u^2 - u^4)\,du
Step 3: Integrate term by term.
=−u33+u55+C= -\frac{u^3}{3} + \frac{u^5}{5} + C
Step 4: Substitute back u = cos x.
=−cos⁡3x3+cos⁡5x5+C= -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C
Answer: −cos⁡3x3+cos⁡5x5+C\displaystyle -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C

Another Example

Problem: Evaluate ∫cos⁡2x dx\int \cos^2 x\,dx.
Step 1: Apply the power-reduction identity cos²x = (1 + cos 2x)/2.
∫cos⁡2x dx=∫1+cos⁡2x2 dx\int \cos^2 x\,dx = \int \frac{1 + \cos 2x}{2}\,dx
Step 2: Integrate each term separately.
=x2+sin⁡2x4+C= \frac{x}{2} + \frac{\sin 2x}{4} + C
Answer: x2+sin⁡2x4+C\displaystyle \frac{x}{2} + \frac{\sin 2x}{4} + C

Why It Matters

Trig integrals appear throughout AP Calculus AB/BC and university-level Calculus II, where they form a core technique category alongside integration by parts and partial fractions. In physics and engineering, they arise when computing work done by oscillating forces, analyzing AC circuits, and finding Fourier coefficients. Mastering the identity-based strategies here saves significant time on exams and in applied problem-solving.

Common Mistakes

Mistake: Forgetting the negative sign when integrating sin x (writing cos x + C instead of −cos x + C).
Correction: Remember that the derivative of cos x is −sin x, so ∫sin⁡x dx=−cos⁡x+C\int \sin x\,dx = -\cos x + C. Always verify by differentiating your answer.
Mistake: Using the power-reduction identity when the exponent is odd, leading to unnecessarily complex expressions.
Correction: When one exponent is odd, the peel-and-substitute method is simpler and faster. Reserve the half-angle identities for cases where both exponents are even.

Related Terms