List of Integrals of Trigonometric Functions: Full Table

What Is The List Of Integrals Of Trigonometric Functions?

The list of integrals of trigonometric functions is a reference table of antiderivatives for integrands built from $\sin x$, $\cos x$, $\tan x$, $\cot x$, $\sec x$ and $\csc x$, alone, raised to powers, multiplied together or placed in a denominator. Each row gives a function $F(x)$ whose derivative is the integrand, so every entry can be checked by differentiating it back. Its anchor rows are

$$\int \sin x,dx = -\cos x + C, \qquad \int \cos x,dx = \sin x + C,$$

where $C$ is the constant of integration, present on every indefinite row. An "integral" here always means an antiderivative unless limits are shown.

This page is the lookup table. The methods that produce the rows (u-substitution for odd powers, power-reduction for even powers, the secant derivation) are taught step by step on the companion page on integration of trigonometric functions, so the notes here stay short.

The symbols used in every table:

How Do You Read A Table Of Trigonometric Integrals?

You read each row as "the derivative of the result is the integrand, on any interval where both are defined". Three habits keep table lookups honest.

• Radians only. Every row assumes $x$ is measured in radians. In degrees, $\frac{d}{dx}\sin x$ picks up a factor of $\frac{\pi}{180}$ and every row changes.

• Absolute values stay. Rows such as $\ln\lvert\cos x\rvert$ keep the bars because $\cos x$ is negative on half its domain.

• Equivalent forms are common. $\ln\lvert\sec x\rvert$ and $-\ln\lvert\cos x\rvert$ are the same function, and $\frac{\sin^2 x}{2}$ and $-\frac{\cos 2x}{4}$ differ only by the constant $\frac{1}{4}$. Two tables can disagree on the look of a row and both be right.

What Are The Integrals Of The Six Basic Trigonometric Functions?

The six basic functions each have a one-line antiderivative. Sine and cosine reverse a derivative directly; the other four need a logarithm.

Four more rows come straight from reading the derivative table backwards:

A Quora user asked, "what is the integral of sec(x) tan(x) and why?" The answer is $\sec x + C$, and the "why" is the last table: the derivative of $\sec x = (\cos x)^{-1}$ is $-(\cos x)^{-2}\cdot(-\sin x) = \frac{\sin x}{\cos^2 x}$, which is exactly $\sec x\tan x$. Reading a derivative backwards is the oldest way to fill an integral table.

What Happens When The Angle Is A Linear Expression?

When the angle is $ax + b$ instead of $x$, the result keeps its shape and gains a factor of $\frac{1}{a}$. That factor undoes the chain rule, which multiplies by $a$ when you differentiate.

A Brainly question asks to "Evaluate $\int \tan(2x),dx$ for $-\frac{\pi}{2} < x < \frac{\pi}{2}$". With $a = 2$, $b = 0$, the row gives $-\frac{1}{2}\ln\lvert\cos 2x\rvert + C$. The stated interval matters: $\cos 2x = 0$ at $x = \pm\frac{\pi}{4}$, so the formula holds on each piece between those points, not across them.

What Are The Integrals Of Squares Of Trigonometric Functions?

Squares need an identity before they integrate. Sine and cosine squared use the power-reduction forms of the double-angle formula, $\sin^2 x = \frac{1 - \cos 2x}{2}$ and $\cos^2 x = \frac{1 + \cos 2x}{2}$; tangent and cotangent squared use the Pythagorean identities $\tan^2 x = \sec^2 x - 1$ and $\cot^2 x = \csc^2 x - 1$.

Look at the first two rows together. Their sum is $x + C$, which matches $\sin^2 x + \cos^2 x = 1$, and that one-line consistency check catches most sign slips in this table.

What Are The Integrals Of Higher Powers Of Trigonometric Functions?

Higher powers follow two patterns. Odd powers of sine or cosine split off one factor and substitute; even powers reduce with the double-angle identity again, and powers of tangent and secant borrow $\tan^2 x = \sec^2 x - 1$.

The $\sec^3 x$ row is the one textbooks single out, and it appears verbatim in a Shaalaa question bank: "Evaluate the following: $\int \sec^3 x,dx$". It needs integration by parts with $u = \sec x$ and $dv = \sec^2 x,dx$:

$$\int \sec^3 x,dx = \sec x\tan x - \int \sec x\tan^2 x,dx$$

$$= \sec x\tan x - \int \sec^3 x,dx + \int \sec x,dx$$

The unknown integral appears on both sides, so move it left and divide by $2$:

$$\int \sec^3 x,dx = \frac{1}{2}\big(\sec x\tan x + \ln\lvert\sec x + \tan x\rvert\big) + C$$

What Are The Reduction Formulas For Trigonometric Integrals?

A reduction formula writes the integral of an $n$th power in terms of the integral of a lower power, so any power can be reached by repeating it. Each one comes from integration by parts, and each can be checked by differentiating its boundary term.

Each formula steps down by $2$, so it ends at a power of $0$ or $1$, both of which sit in the basic tables. The $\sec^n$ row with $n = 3$ reproduces the $\sec^3 x$ result above in one line, which is a quick way to trust the general formula.

On $\left[0, \frac{\pi}{2}\right]$ the boundary term vanishes, and the sine and cosine formulas collapse into Wallis's formula:

$$\int_{0}^{\pi/2} \sin^n x,dx = \int_{0}^{\pi/2} \cos^n x,dx = \frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdots \times \begin{cases} \frac{\pi}{2} & n \text{ even} \ 1 & n \text{ odd} \end{cases}$$

So $\int_{0}^{\pi/2} \sin^4 x,dx = \frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} = \frac{3\pi}{16}$ and $\int_{0}^{\pi/2} \sin^5 x,dx = \frac{4}{5}\cdot\frac{2}{3} = \frac{8}{15}$.

What Are The Integrals Of Products Of Trigonometric Functions?

Products split into three families: powers of sine times powers of cosine, products of sines and cosines with different frequencies, and powers of tangent times powers of secant. A Brainly question puts the first family plainly: "Which identity is most useful in evaluating $\int \sin^3 x\cos^2 x,dx$?" The answer is $\sin^2 x = 1 - \cos^2 x$, applied after saving one $\sin x$ for $du$. Then integration by substitution with $u = \cos x$ gives $-\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C$.

That choice generalises into a decision table for $\int \sin^m x\cos^n x,dx$:

Standard product rows:

The three frequency rows require $m^2 \neq n^2$; when $m = n$, use the square rows instead. They come from the product-to-sum identities, such as $\sin A\cos B = \frac{1}{2}\big[\sin(A-B) + \sin(A+B)\big]$, and they are the reason Fourier series work.

For $\int \tan^m x\sec^n x,dx$ the rule mirrors the sine-cosine table: with $n$ even, save $\sec^2 x$ and set $u = \tan x$; with $m$ odd, save $\sec x\tan x$ and set $u = \sec x$.

How Do You Integrate $\frac{1}{a + b\sin x}$ And $\frac{1}{a + b\cos x}$?

These denominators fall to the Weierstrass substitution $t = \tan\frac{x}{2}$, which turns any rational function of $\sin x$ and $\cos x$ into a rational function of $t$. It rests on the half-angle formulas:

$$\sin x = \frac{2t}{1+t^2}, \qquad \cos x = \frac{1-t^2}{1+t^2}, \qquad dx = \frac{2,dt}{1+t^2}$$

After the swap, the integral becomes $\int \frac{2,dt}{(a+b) + (a-b)t^2}$ for the cosine case, and the sign of $a^2 - b^2$ decides whether the answer is an arctangent or a logarithm. This is the row that takes longest to trust: the case split looks arbitrary until you see it is the discriminant of a quadratic in $t$.

For a negative $b$ in the cosine rows, factor out $-1$ first. Every one of these antiderivatives contains $\tan\frac{x}{2}$, which jumps at $x = \pi$, so each formula holds on one interval between those jumps; Mistake 4 below shows what happens when a definite integral crosses one.

What Are The Integrals Of Trigonometric Functions Multiplied By Other Functions?

When a trig function multiplies a power of $x$ or an exponential, integration by parts produces the result; for exponentials it runs twice and the original integral reappears, the same trick as the $\sec^3 x$ row.

The last two rows connect this list to integration of exponential functions: they model damped oscillations, a spring or a circuit whose swing shrinks by a factor of $e^{ax}$ with $a < 0$.

Which Trigonometric Integrals Have No Elementary Formula?

Some short-looking trigonometric integrands have no antiderivative made of finitely many elementary functions. Their integrals exist, but they are named as new special functions instead of formulas.

The NIST Digital Library of Mathematical Functions defines $\operatorname{Si}(z) = \int_{0}^{z} \frac{\sin t}{t},dt$ (equation 6.2.9) and $\operatorname{Ci}(z) = -\int_{z}^{\infty} \frac{\cos t}{t},dt$ (equation 6.2.11). The proof that no elementary formula exists comes from Liouville's theory of integration in finite terms, which sits well beyond a first calculus course.

What Are The Standard Definite Integrals Of Trigonometric Functions?

Many definite trig integrals have exact values even when the indefinite form is long. Symmetry and periodicity do most of the work, including the property $\int_{0}^{a} f(x),dx = \int_{0}^{a} f(a-x),dx$ from the properties of definite integrals, which shows at once that sine and cosine powers share the same value on $\left[0, \frac{\pi}{2}\right]$.

The last row is an improper integral known as the Dirichlet integral; it equals $\lim_{x \to \infty}\operatorname{Si}(x)$. The two orthogonality rows are the reason a sound wave can be split into separate frequencies: integrating against $\sin nx$ picks out one frequency and sends every other one to zero.

Examples Of Integrals Of Trigonometric Functions

Example 1

Use the list to find $\int \big(4\cos 3x - \sec^2 x\big),dx$.

Split the integral and use the linear-argument row with $a = 3$:

$$\int 4\cos 3x,dx = 4\cdot\frac{1}{3}\sin 3x = \frac{4}{3}\sin 3x$$

$$\int \sec^2 x,dx = \tan x$$

$$\int \big(4\cos 3x - \sec^2 x\big),dx = \frac{4}{3}\sin 3x - \tan x + C$$

Final answer: $\frac{4}{3}\sin 3x - \tan x + C$

Example 2

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

The power rule looks tempting, treating $\sin x$ like $x$:

$$\text{Attempt: } \int \sin^2 x,dx = \frac{\sin^3 x}{3} + C$$

Differentiate the attempt to check it. The chain rule gives

$$\frac{d}{dx}\left[\frac{\sin^3 x}{3}\right] = \sin^2 x\cos x$$

That has an extra factor of $\cos x$, so the attempt is wrong. The power rule only reverses when the derivative of the inside is already present. Use power reduction instead:

$$\sin^2 x = \frac{1 - \cos 2x}{2}$$

$$\int \sin^2 x,dx = \int \frac{1}{2},dx - \int \frac{\cos 2x}{2},dx$$

$$= \frac{x}{2} - \frac{\sin 2x}{4} + C$$

Check: $\frac{d}{dx}\left[\frac{x}{2} - \frac{\sin 2x}{4}\right] = \frac{1}{2} - \frac{\cos 2x}{2} = \sin^2 x$.

Final answer: $\frac{x}{2} - \frac{\sin 2x}{4} + C$

Example 3

Find $\int \sin^3 x\cos^4 x,dx$.

The sine power is odd, so save one $\sin x$ and convert $\sin^2 x = 1 - \cos^2 x$. Let $u = \cos x$, so $du = -\sin x,dx$:

$$\int \sin^3 x\cos^4 x,dx = \int \big(1 - \cos^2 x\big)\cos^4 x\sin x,dx$$

$$= -\int \big(1 - u^2\big)u^4,du$$

$$= -\frac{u^5}{5} + \frac{u^7}{7} + C$$

$$= -\frac{\cos^5 x}{5} + \frac{\cos^7 x}{7} + C$$

The step that looks skippable is saving the single $\sin x$ before converting; convert all three powers and no $du$ is left to absorb.

Final answer: $-\frac{\cos^5 x}{5} + \frac{\cos^7 x}{7} + C$

Example 4

Evaluate $\int_{0}^{\pi/2} \sin^6 x,dx$.

Use Wallis's formula with $n = 6$ (even, so the last factor is $\frac{\pi}{2}$):

$$\int_{0}^{\pi/2} \sin^6 x,dx = \frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}$$

$$= \frac{15\pi}{96} = \frac{5\pi}{32}$$

Final answer: $\frac{5\pi}{32} \approx 0.4909$

Example 5

Evaluate $\int_{0}^{\pi/2} \frac{dx}{5 + 3\cos x}$.

Here $a = 5$, $b = 3$ and $a^2 > b^2$, so use the arctangent row with $\sqrt{a^2 - b^2} = 4$ and $\sqrt{\frac{a-b}{a+b}} = \sqrt{\frac{2}{8}} = \frac{1}{2}$:

$$\int \frac{dx}{5 + 3\cos x} = \frac{2}{4}\arctan\left(\frac{1}{2}\tan\frac{x}{2}\right) + C$$