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Concept Version 10
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Trigonometric Integrals

The trigonometric integrals are a specific set of functions used to simplify complex mathematical expressions in order to evaluate them.

Learning Objective

  • Solve basic trigonometric integrals


Key Points

    • Some of the expressions for the trigonometric integrals are found using properties of trigonometric functions.
    • Some of the expressions were derived using techniques like integration by parts.
    • There is no guarantee that a trigonometric integral has an analytic expression.

Terms

  • trigonometric

    relating to the functions used in trigonometry: $\sin$, $\cos$, $\tan$, $\csc$, $\cot$, $\sec$

  • integral

    also sometimes called antiderivative; the limit of the sums computed in a process in which the domain of a function is divided into small subsets and a possibly nominal value of the function on each subset is multiplied by the measure of that subset, all these products then being summed


Full Text

Trigonometric Integrals

The trigonometric integrals are a family of integrals which involve trigonometric functions ($\sin$, $\cos$, $\tan$, $\csc$, $\cot$, $\sec$). The following is a list of integrals of trigonometric functions. Some of them were computed using properties of the trigonometric functions, while others used techniques such as integration by parts. 

Generally, if the function, $\sin(x)$, is any trigonometric function, and $\cos(x)$ is its derivative, then

$\displaystyle{\int a\cos nx\;\mathrm{d}x = \frac{a}{n}\sin nx+C}$

In all formulas, the constant $a$ is assumed to be nonzero, while $C$ denotes the integration constant.

Integrands Involving Only Sine:

$\displaystyle{\int\sin ax\;\mathrm{d}x = -\frac{1}{a}\cos ax+C\,\! \\ \int\sin^2 {ax}\;\mathrm{d}x = \frac{x}{2} - \frac{1}{4a} \sin 2ax +C= \frac{x}{2} - \frac{1}{2a} \sin ax\cos ax +C\! \\ \int\sin^3 {ax}\;\mathrm{d}x = \frac{\cos 3ax}{12a} - \frac{3 \cos ax}{4a} +C\! \\ \int x\sin^2 {ax}\;\mathrm{d}x = \frac{x^2}{4} - \frac{x}{4a} \sin 2ax - \frac{1}{8a^2} \cos 2ax +C\! \\ \int x^2\sin^2 {ax}\;\mathrm{d}x = \frac{x^3}{6} - \left( \frac {x^2}{4a} - \frac{1}{8a^3} \right) \sin 2ax - \frac{x}{4a^2} \cos 2ax +C\!}$

Integrands Involving Only Cosine:

$\int\cos ax\;\mathrm{d}x = \frac{1}{a}\sin ax+C$

$\int\cos^2 {ax}\;\mathrm{d}x = \frac{x}{2} + \frac{1}{4a} \sin 2ax +C = \frac{x}{2} + \frac{1}{2a} \sin ax\cos ax +C$

$\int\cos^n ax\;\mathrm{d}x = \frac{\cos^{n-1} ax\sin ax}{na} + \frac{n-1}{n}\int\cos^{n-2} ax\;\mathrm{d}x \qquad\mbox{(for }n>0\mbox{)}$

$\int x\cos ax\;\mathrm{d}x = \frac{\cos ax}{a^2} + \frac{x\sin ax}{a}+C$

$\int x^2\cos^2 {ax}\;\mathrm{d}x = \frac{x^3}{6} + \left( \frac {x^2}{4a} - \frac{1}{8a^3} \right) \sin 2ax + \frac{x}{4a^2} \cos 2ax +C$

Integrands Involving Only Tangent:

$\begin{aligned}\int\tan ax\;\mathrm{d}x &= -\frac{1}{a}\ln \left|\cos ax \right|+C\\ & = \frac{1}{a}\ln \left|\sec ax \right|+C\end{aligned}$

$\displaystyle{\int\tan^n ax\;\mathrm{d}x = \frac{1}{a(n-1)}\tan^{n-1} ax-\int\tan^{n-2} ax\;\mathrm{d}x}$

where $n \neq 1$; and

$\displaystyle{\int\frac{\mathrm{d}x}{q \tan ax + p} = \frac{1}{p^2 + q^2}(px + \frac{q}{a}\ln \left|q\sin ax + p\cos ax \right|)+C }$

where $p^2 + q^2 \neq 0$.

Integrands Involving Only Secant:

$\displaystyle{\int \sec{ax} \, \mathrm{d}x = \frac{1}{a}\ln{\left| \sec{ax} + \tan{ax}\right|}+C}$

$\displaystyle{\int \sec^2{x} \, \mathrm{d}x = \tan{x}+C}$

Integrands involving only cosecant:

$\displaystyle{\int \csc{ax} \, \mathrm{d}x = \frac{1}{a}\ln{\left| \csc{ax}-\cot{ax}\right|}+C}$

$\displaystyle{\int \csc^2{x} \, \mathrm{d}x = -\cot{x}+C}$

Complicated Trigonometric Integrals

We now look at integrals involving the product of a power of $\sin x$ and a power of $\cos x$. Two simple examples of such integrals are $\int \sin^k x \cos x \; \mathrm d x$ and $\int \cos^k x \sin x\; \mathrm d x$ , which can be solved used the substitutions $u = \sin x$ and $u = \cos x$ , respectively. We now consider the more general case of $\int \sin^n x \cos^m x\; \mathrm d x$ , where $n$ and $m$ are positive integers.

  • If $n$ is odd, we can pull out one factor of $\sin x$ , convert the rest to cosines using the identity $\sin^2 x + \cos^2 x = 1$ , and then use the substitution $u = \cos x$ .
  • If $m$ is odd, we can pull out one factor of $\cos x$ , convert the rest to sines using the identity $\sin^2 x + \cos^2 x = 1$ , and then use the substitution $u = \sin x$.
  • If both $n$ and $m$ are odd, we can use either of the above two methods.
  • If both $n$ and $m$ are even, then we can try to use a combination of the following three identities: $\cos^2 x = \frac{1}{2} (1 + \cos 2x)$, $\sin^2 x = \frac{1}{2} (1 - \sin 2x)$, and $\sin x \cos x = \frac{1}{2} \sin 2x$.
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