Trigonometric Integrals: Methods & Examples

Trigonometric integrals are integrals of powers and products of sin x, cos x, tan x and sec x. If sine or cosine has an odd power, save one factor, change the rest with sin²x + cos²x = 1, and substitute. If both powers are even, use the half-angle identities to lower the powers first.

Before this lesson: U-Substitution, Fundamental Trig Identities and Half-Angle Formulas

The main idea

You can’t integrate sin³x with the power rule, because the power rule needs the derivative of sin x, which is cos x, to be there. The trick is to make a factor of the derivative appear.

Write sin³x as sin²x · sin x and change sin²x into 1 − cos²x. Now every factor is a cosine except one sin x, and that sin x dx is (up to a sign) du for u = cos x. An odd power always leaves one factor to save. Even powers don’t, so you use identities that lower the power instead.

Decision chart for the integral of sin to the m times cos to the n. If the power of sine is odd: save one sin x, change the rest to cosines, and let u = cos x. If the power of cosine is odd: save one cos x, change the rest to sines, and let u = sin x. If both powers are even: use the half-angle identities. — Pick the strategy from the powers: an odd power lets you save one factor and substitute; even powers need half-angle identities.

Trigonometric integral strategies

Integral	Strategy	Substitution
$\int \sin^m x\cos^n x\,dx$ , m odd	Save one sin x; use $\sin^2 x = 1 - \cos^2 x$	u = cos x
$\int \sin^m x\cos^n x\,dx$ , n odd	Save one cos x; use $\cos^2 x = 1 - \sin^2 x$	u = sin x
$\int \sin^m x\cos^n x\,dx$ , both even	Use the half-angle identities below	Often none needed
$\int \tan^m x\sec^n x\,dx$ , n even	Save sec²x; use $\sec^2 x = 1 + \tan^2 x$	u = tan x
$\int \tan^m x\sec^n x\,dx$ , m odd	Save sec x tan x; use $\tan^2 x = \sec^2 x - 1$	u = sec x

The half-angle identities:

\sin^2 x = \dfrac{1 - \cos 2x}{2}, \qquad \cos^2 x = \dfrac{1 + \cos 2x}{2}

Trigonometric integral examples

Example 1: An odd power of sine

Find $\int \sin^3 x\,dx$ .

\sin^3 x = \sin^2 x\cdot\sin x = (1 - \cos^2 x)\sin x

Save one sin x and use the Pythagorean identity (see fundamental trig identities).

u = \cos x

Let u be cosine.

du = -\sin x\,dx

Differentiate u, so sin x dx = −du.

\int \sin^3 x\,dx = \int -(1 - u^2)\,du

Rewrite in u (see u-substitution).

= -u + \dfrac{u^3}{3} + C

Integrate each term.

= -\cos x + \dfrac{\cos^3 x}{3} + C

Substitute back.

Answer: $-\cos x + \dfrac{\cos^3 x}{3} + C$

Example 2: An odd power of cosine

Find $\int \sin^2 x\cos^3 x\,dx$ .

\cos^3 x = (1 - \sin^2 x)\cos x

The cosine power is odd: save one cos x.

u = \sin x

Let u be sine.

du = \cos x\,dx

Differentiate u.

\int \sin^2 x\cos^3 x\,dx = \int u^2(1 - u^2)\,du

Rewrite in u.

= \int (u^2 - u^4)\,du

Distribute (see distributive property).

= \dfrac{u^3}{3} - \dfrac{u^5}{5} + C

Integrate each term.

= \dfrac{\sin^3 x}{3} - \dfrac{\sin^5 x}{5} + C

Substitute back.

Answer: $\dfrac{\sin^3 x}{3} - \dfrac{\sin^5 x}{5} + C$

Example 3: An even power

Find $\int \cos^2 x\,dx$ .

\cos^2 x = \dfrac{1 + \cos 2x}{2}

The power is even: use the half-angle identity (see half-angle formulas).

\int \cos^2 x\,dx = \int \left(\dfrac{1}{2} + \dfrac{1}{2}\cos 2x\right)dx

Split the fraction.

= \dfrac{x}{2} + \dfrac{1}{2}\cdot\dfrac{\sin 2x}{2} + C

Integrate; cos 2x integrates to (sin 2x)/2.

= \dfrac{x}{2} + \dfrac{\sin 2x}{4} + C

Multiply.

Answer: $\dfrac{x}{2} + \dfrac{\sin 2x}{4} + C$

Example 4: Secant and tangent

Find $\int \sec^4 x\,dx$ .

\sec^4 x = \sec^2 x\cdot\sec^2 x = (1 + \tan^2 x)\sec^2 x

The secant power is even: save sec²x and use sec²x = 1 + tan²x.

u = \tan x

Let u be tangent.

du = \sec^2 x\,dx

Differentiate u (see derivatives of trig functions).

\int \sec^4 x\,dx = \int (1 + u^2)\,du

Rewrite in u.

= u + \dfrac{u^3}{3} + C = \tan x + \dfrac{\tan^3 x}{3} + C

Integrate and substitute back.

Answer: $\tan x + \dfrac{\tan^3 x}{3} + C$

Example 5: Average power in a circuit

In an AC circuit, the power used by a heater is $P(t) = 100\sin^2(120\pi t)$ watts, where t is in seconds. One cycle lasts 1/60 of a second. Find the average power over one cycle.

P_{\text{avg}} = 60\int_0^{1/60} 100\sin^2(120\pi t)\,dt

Average value: divide by the length 1/60, which is the same as multiplying by 60 (see average value of a function).

\sin^2(120\pi t) = \dfrac{1 - \cos(240\pi t)}{2}

Half-angle identity with x = 120πt.

P_{\text{avg}} = 6000\int_0^{1/60} \dfrac{1 - \cos(240\pi t)}{2}\,dt

Replace the integrand; 60 · 100 = 6000.

= 3000\Big[t - \dfrac{\sin(240\pi t)}{240\pi}\Big]_0^{1/60}

Factor out 1/2 and integrate.

= 3000\left(\dfrac{1}{60} - 0\right) = 50

sin(4π) = 0 and sin 0 = 0.

Answer: the average power is 50 watts, exactly half the peak power of 100 watts. That is why the average of sin² over a full cycle is 1/2.

Common mistakes

Mistake	Example of the error	Fix
Using the power rule on a trig power	$\int \sin^2 x\,dx = \dfrac{\sin^3 x}{3} + C$	There’s no cos x factor. Use the half-angle identity: $\dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$ .
Saving the wrong factor	In $\int \sin^3 x\,dx$ , letting u = sin x	With an odd power of sine, save sin x and let u = cos x.
Mixing up the half-angle signs	$\sin^2 x = \dfrac{1 + \cos 2x}{2}$	Sine gets the minus: $\sin^2 x = \dfrac{1 - \cos 2x}{2}$ .
Forgetting the 1/2 from cos 2x	$\int \cos 2x\,dx = \sin 2x + C$	u = 2x gives du = 2 dx: $\dfrac{\sin 2x}{2} + C$ .
Dropping the sign of du	$\int \sin x\cos^2 x\,dx = \dfrac{\cos^3 x}{3} + C$	du = −sin x dx, so the answer is $-\dfrac{\cos^3 x}{3} + C$ .

Related methods

Products with different angles

Integrals like $\int \sin 3x\cos 2x\,dx$ use the product-to-sum identity $\sin A\cos B = \dfrac{1}{2}\big(\sin(A - B) + \sin(A + B)\big)$ . Then each piece is a basic integral.

Odd powers of secant alone

$\int \sec^3 x\,dx$ needs integration by parts (see integration by parts) and gives $\dfrac{1}{2}\big(\sec x\tan x + \ln|\sec x + \tan x|\big) + C$ .

Trig substitution

These integrals show up again in the next lesson, Trigonometric Substitution, where a root like $\sqrt{1 - x^2}$ turns into a trig integral.

Practice problems

Find $\int \cos^3 x\,dx$ .
Find $\int \sin x\cos^2 x\,dx$ .
Find $\int \sin^3 x\cos^2 x\,dx$ .
Find $\int \sin^2 x\,dx$ .
Evaluate $\int_0^{\pi/2} \cos^2 x\,dx$ .
Find $\int \sin^5 x\,dx$ .
Find $\int \tan x\sec^2 x\,dx$ .
Find $\int \tan^3 x\sec x\,dx$ .
Find $\int \tan^2 x\,dx$ .
Find $\int \sin^2 x\cos^2 x\,dx$ .

Show answers

\cos^3 x = (1 - \sin^2 x)\cos x

Save one cos x.

\int \cos^3 x\,dx = \int (1 - u^2)\,du

Let u = sin x, du = cos x dx.

= u - \dfrac{u^3}{3} + C = \sin x - \dfrac{\sin^3 x}{3} + C

Integrate and substitute back.

Answer: $\sin x - \dfrac{\sin^3 x}{3} + C$

\int \sin x\cos^2 x\,dx = \int -u^2\,du

Let u = cos x, du = −sin x dx.

= -\dfrac{u^3}{3} + C = -\dfrac{\cos^3 x}{3} + C

Integrate and substitute back.

Answer: $-\dfrac{\cos^3 x}{3} + C$

\sin^3 x\cos^2 x = (1 - \cos^2 x)\cos^2 x\sin x

Save one sin x.

\int \sin^3 x\cos^2 x\,dx = \int -(u^2 - u^4)\,du

Let u = cos x, du = −sin x dx.

= -\dfrac{u^3}{3} + \dfrac{u^5}{5} + C = -\dfrac{\cos^3 x}{3} + \dfrac{\cos^5 x}{5} + C

Integrate and substitute back.

Answer: $-\dfrac{\cos^3 x}{3} + \dfrac{\cos^5 x}{5} + C$

\int \sin^2 x\,dx = \int \left(\dfrac{1}{2} - \dfrac{1}{2}\cos 2x\right)dx

Half-angle identity.

= \dfrac{x}{2} - \dfrac{\sin 2x}{4} + C

Integrate each term.

Answer: $\dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$

\int_0^{\pi/2} \cos^2 x\,dx = \Big[\dfrac{x}{2} + \dfrac{\sin 2x}{4}\Big]_0^{\pi/2}

Use Example 3.

= \left(\dfrac{\pi}{4} + \dfrac{\sin\pi}{4}\right) - 0 = \dfrac{\pi}{4}

sin π = 0.

Answer: $\dfrac{\pi}{4}$

\sin^5 x = (1 - \cos^2 x)^2\sin x

Save one sin x; sin⁴x = (sin²x)².

\int \sin^5 x\,dx = \int -(1 - 2u^2 + u^4)\,du

Let u = cos x, du = −sin x dx; expand the square.

= -u + \dfrac{2u^3}{3} - \dfrac{u^5}{5} + C

Integrate each term.

= -\cos x + \dfrac{2\cos^3 x}{3} - \dfrac{\cos^5 x}{5} + C

Substitute back.

Answer: $-\cos x + \dfrac{2\cos^3 x}{3} - \dfrac{\cos^5 x}{5} + C$

\int \tan x\sec^2 x\,dx = \int u\,du = \dfrac{u^2}{2} + C

Let u = tan x, du = sec²x dx.

= \dfrac{\tan^2 x}{2} + C

Substitute back.

Answer: $\dfrac{\tan^2 x}{2} + C$

\tan^3 x\sec x = (\sec^2 x - 1)\sec x\tan x

Save sec x tan x; use tan²x = sec²x − 1.

\int \tan^3 x\sec x\,dx = \int (u^2 - 1)\,du

Let u = sec x, du = sec x tan x dx.

= \dfrac{u^3}{3} - u + C = \dfrac{\sec^3 x}{3} - \sec x + C

Integrate and substitute back.

Answer: $\dfrac{\sec^3 x}{3} - \sec x + C$

\int \tan^2 x\,dx = \int (\sec^2 x - 1)\,dx

Use tan²x = sec²x − 1.

= \tan x - x + C

Integrate each term.

Answer: $\tan x - x + C$

10.

\sin^2 x\cos^2 x = \dfrac{1}{4}\sin^2 2x

sin x cos x = (sin 2x)/2, so square both sides (see double-angle formulas).

\dfrac{1}{4}\sin^2 2x = \dfrac{1}{8}(1 - \cos 4x)

Half-angle identity with angle 2x.

\int \dfrac{1}{8}(1 - \cos 4x)\,dx = \dfrac{x}{8} - \dfrac{\sin 4x}{32} + C

Integrate each term.

Answer: $\dfrac{x}{8} - \dfrac{\sin 4x}{32} + C$

Trigonometric integrals FAQ

How do you integrate sin²x?

Use the half-angle identity sin²x = (1 − cos 2x)/2. Then $\int \sin^2 x\,dx = \dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$ .

When do you use half-angle identities in integrals?

Use them when sine and cosine both have even powers (including 0). Odd powers are easier: save one factor and substitute.

What is the integral of sec²x?

It is tan x + C, because the derivative of tan x is sec²x.

What is the integral of tan²x?

Write tan²x = sec²x − 1. The integral is tan x − x + C.

Summary

Odd power of sine: save sin x, convert to cosines, u = cos x.
Odd power of cosine: save cos x, convert to sines, u = sin x.
Both even: use sin²x = (1 − cos 2x)/2 and cos²x = (1 + cos 2x)/2.
Even power of secant: save sec²x, u = tan x.
Odd power of tangent with a secant: save sec x tan x, u = sec x.

Previous lesson: Integration by Parts

Next lesson: Trigonometric Substitution

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