[CALCULUS] Trigonometric integrals

This page integrates powers and products of sine, cosine, tangent and secant. Each case is recognised by the shape of the integrand:

One power odd in $\int \cos^m x \sin^n x \, dx$ : peel one factor and substitute $u=\sin x$ or $u=\cos x$ .
Both powers even: use the double-angle identities to lower the powers.
Different frequencies, such as $\int \cos(mx)\sin(nx) \, dx$ : use a product-to-sum identity.
Powers of $\tan x$ and $\sec x$ : use $1+\tan^2 x=\sec^2 x$ with $\frac{d}{dx}\tan x=\sec^2 x$ and $\frac{d}{dx}\sec x=\tan x\sec x$ .

For high powers that recur with a lower index, reduction formulae organise the repeated integration by parts. A quadratic surd calls for trigonometric and hyperbolic substitutions instead.

Throughout, $m$ and $n$ are non-negative integers. An indefinite antiderivative involving $\tan x$ or $\sec x$ holds on an interval that does not cross a zero of $\cos x$ .

Odd powers of sine or cosine

[METHOD]

Odd power of sine or cosine

To evaluate $\int \cos^m x \sin^n x \, dx$ when at least one of $m,n$ is odd:

If $m$ is odd, factor out $\cos x$ to obtain $\int \cos^{m-1} x \sin^n x \cos x \, dx$ . If $n$ is odd, factor out $\sin x$ to obtain $\int \cos^m x \sin^{n-1} x \sin x \, dx$ .
The leftover power is even. Replace it using $\sin^2 x+\cos^2 x=1$ .
Substitute $u=\sin x$ or $u=\cos x$ . The binomial expansion $(a+b)^k$ is usually needed before the powers of $u$ can be integrated.

Always peel from an odd power. Peeling from an even power leaves an odd remainder, which $1=\sin^2 x+\cos^2 x$ cannot turn into a polynomial in the substitution variable. If both $m$ and $n$ are odd, either factor works, although the two antiderivatives need not look the same.

[EXAMPLE]

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

Here $m$ is even and $n$ is odd. Peel $\sin x$ and replace $\sin^2 x=1-\cos^2 x$ :

\begin{aligned} \int \cos^4 x \sin^3 x \, dx &=\int \cos^4 x \sin^2 x \sin x \, dx \\ &=\int \cos^4 x (1-\cos^2 x) \sin x \, dx \\ &=\int (\cos^4 x-\cos^6 x) \sin x \, dx. \end{aligned}

Let $u=\cos x$ , so $du=-\sin x \, dx$ . The integral is

-\int (u^4-u^6) \, du=-\frac{u^5}{5}+\frac{u^7}{7}+C=-\frac{\cos^5 x}{5}+\frac{\cos^7 x}{7}+C.

[routine]

[1]:Odd power of sine or cosine

[EXAMPLE]

$\int \cos^5 x \sin^4 x \, dx$

Here $m$ is odd and $n$ is even. Peel $\cos x$ and replace $\cos^2 x=1-\sin^2 x$ :

\begin{aligned} \int \cos^5 x \sin^4 x \, dx &=\int \cos^4 x \sin^4 x \cos x \, dx \\ &=\int (1-\sin^2 x)^2 \sin^4 x \cos x \, dx \\ &=\int \bigl(\sin^4 x-2\sin^6 x+\sin^8 x\bigr) \cos x \, dx. \end{aligned}

Let $u=\sin x$ , so $du=\cos x \, dx$ . The integral is

\int (u^4-2u^6+u^8) \, du=\frac{u^5}{5}-\frac{2u^7}{7}+\frac{u^9}{9}+C=\frac{\sin^5 x}{5}-\frac{2\sin^7 x}{7}+\frac{\sin^9 x}{9}+C.

[exam]

[EXAMPLE]

$\int \cos^5 x \sin^5 x \, dx$

Both $m$ and $n$ are odd. Factoring out $\cos x$ and substituting $u=\sin x$ (so $du=\cos x \, dx$ ),

\begin{aligned} \int \cos^5 x \sin^5 x \, dx &=\int \cos^4 x \sin^5 x \cos x \, dx \\ &=\int (1-\sin^2 x)^2 \sin^5 x \cos x \, dx \\ &=\int \bigl(\sin^5 x-2\sin^7 x+\sin^9 x\bigr) \cos x \, dx \\ &=\int (u^5-2u^7+u^9) \, du \\ &=\frac{u^6}{6}-\frac{2u^8}{8}+\frac{u^{10}}{10}+C \\ &=\frac{\sin^6 x}{6}-\frac{\sin^8 x}{4}+\frac{\sin^{10} x}{10}+C. \end{aligned}

Factoring out $\sin x$ instead produces a different-looking formula. Both formulas have derivative $\cos^5x\sin^5x$ , so on any interval they differ by a constant.

[exam]

[1]:Odd power of sine or cosine

Even powers of sine and cosine

[METHOD]

Both powers even

To evaluate $\int \cos^m x \sin^n x \, dx$ when both $m$ and $n$ are even, use

\cos^2\alpha=\frac{1+\cos 2\alpha}{2},\qquad \sin^2\alpha=\frac{1-\cos 2\alpha}{2}

to rewrite the integrand as a sum of integrals of the form $\int \cos^k(2x) \, dx$ . Integrate any odd $k$ by the odd-power substitution in $2x$ ; apply the double-angle identities again to any remaining even $k$ .

After expanding, the constant and $\cos 2x$ terms integrate immediately. Because the argument is $2x$ , the odd-power substitution is $u=\sin 2x$ , with $du=2\cos 2x\, dx$ .

[1]:Odd power of sine or cosine
[2]: $\int\sqrt{1-x^2}\,dx$
[3]: $\int \cos^2 x \, dx$ and $\int \sin^2 x \, dx$
[4]: $\int \cos^2 x \sin^4 x \, dx$
[5]:Product-to-sum

[EXAMPLE]

$\int \cos^2 x \, dx$ and $\int \sin^2 x \, dx$

When one of $m$ or $n$ is zero the identities are the whole calculation:

\int \cos^2 x \, dx=\int \frac{1+\cos 2x}{2} \, dx=\frac{x}{2}+\frac{\sin 2x}{4}+C,

\int \sin^2 x \, dx=\int \frac{1-\cos 2x}{2} \, dx=\frac{x}{2}-\frac{\sin 2x}{4}+C.

[routine]

[1]:Both powers even
[2]:A repeated irreducible quadratic

[EXAMPLE]

$\int \cos^2 x \sin^4 x \, dx$

Both powers are even, so

\int \cos^2 x \sin^4 x \, dx=\int \frac{1+\cos 2x}{2}\left(\frac{1-\cos 2x}{2}\right)^2 \, dx=\frac{1}{8}\int (1+\cos 2x)(1-\cos 2x)^2 \, dx.

Expanding and cancelling $-\cos 2x$ against $+\cos 2x$ gives

(1+\cos 2x)(1-2\cos 2x+\cos^2 2x)=1-\cos 2x-\cos^2 2x+\cos^3 2x,

so the integral is

\frac{1}{8}\left(\int 1 \, dx-\int\cos 2x \, dx-\int\cos^2 2x \, dx+\int\cos^3 2x \, dx\right).

The first two integrals evaluate to $x$ and $\frac{1}{2}\sin 2x$ . For the even square, $\cos^2 2x=\frac{1+\cos 4x}{2}$ . For the odd cube, $\cos^3 2x=(1-\sin^2 2x)\cos 2x$ . Thus the integral equals

\frac{1}{8}\left[x-\frac{\sin 2x}{2}-\int\frac{1+\cos 4x}{2} \, dx+\int(1-\sin^2 2x)\cos 2x \, dx\right].

The remaining cosine-square integral is $\frac{x}{2}+\frac{\sin 4x}{8}$ . For the last integral, let $u=\sin 2x$ , so $du=2\cos 2x \, dx$ and $\cos 2x \, dx=\tfrac12 \, du$ . Collecting factors,

\frac{x}{8}-\frac{\sin 2x}{16}-\frac{x}{16}-\frac{\sin 4x}{64}+\frac{1}{16}\int (1-u^2) \, du =\frac{x}{16}-\frac{\sin 2x}{16}-\frac{\sin 4x}{64}+\frac{1}{16}\left(u-\frac{u^3}{3}\right)+C.

Back-substituting $u=\sin 2x$ yields

\frac{x}{16}-\frac{\sin 2x}{16}-\frac{\sin 4x}{64}+\frac{\sin 2x}{16}-\frac{\sin^3 2x}{48}+C=\frac{x}{16}-\frac{\sin 4x}{64}-\frac{\sin^3 2x}{48}+C.

[exam]

[1]:Both powers even

Products at distinct frequencies

The methods above treat powers of $\sin x$ and $\cos x$ at a single argument. The integrals $\int \cos(mx)\sin(nx) \, dx$ , $\int \cos(mx)\cos(nx) \, dx$ and $\int \sin(mx)\sin(nx) \, dx$ mix two frequencies and a product-to-sum identity separates them.

[METHOD]

Product-to-sum

Convert the product to a sum, then integrate term by term:

\sin A\cos B=\tfrac12\bigl(\sin(A+B)+\sin(A-B)\bigr)

\cos A\cos B=\tfrac12\bigl(\cos(A+B)+\cos(A-B)\bigr)

\sin A\sin B=-\tfrac12\bigl(\cos(A+B)-\cos(A-B)\bigr)

The minus sign on $\sin A\sin B$ is the one most often dropped.

[1]:Odd power of sine or cosine
[2]:Both powers even
[3]:Average value of a piecewise function
[4]:Seen questions
[5]: $\int \sin 3x \sin 2x \, dx$

[EXAMPLE]

$\int \sin 3x \sin 2x \, dx$

With $A=3x$ and $B=2x$ ,

\begin{aligned} \int \sin 3x \sin 2x \, dx &=-\frac{1}{2}\int \bigl(\cos 5x-\cos x\bigr) \, dx \\ &=-\frac{1}{2}\left(\frac{\sin 5x}{5}-\sin x\right)+C \\ &=-\frac{\sin 5x}{10}+\frac{\sin x}{2}+C. \end{aligned}

[exam]

[1]:Product-to-sum

Powers of tangent and secant

[DEFINITION]

Tangent and secant

The identities used to integrate $\int \tan^m x \sec^n x \, dx$ are

1+\tan^2 x=\sec^2 x,\qquad \frac{d}{dx}\tan x=\sec^2 x,\qquad \frac{d}{dx}\sec x=\tan x\sec x.

The two derivative identities are why $u=\tan x$ and $u=\sec x$ match a leftover factor of $\sec^2 x$ or $\tan x\sec x$ .

[EXAMPLE]

$\int \tan x \, dx$ and $\int \sec x \, dx$

Writing $\tan x=\frac{\sin x}{\cos x}$ ,

\int \tan x \, dx=\int \frac{\sin x}{\cos x} \, dx=-\ln|\cos x|+C.

For the secant integral, multiply numerator and denominator by $\sec x+\tan x$ :

\int \sec x \, dx=\int \frac{\sec x(\sec x+\tan x)}{\sec x+\tan x} \, dx=\int \frac{\sec^2 x+\sec x\tan x}{\sec x+\tan x} \, dx=\ln|\sec x+\tan x|+C,

since the numerator is then the derivative of the denominator.

[routine]

[1]:Tangent and secant

[DEFINITION]

Tangent and secant

The identities used to integrate $\int \tan^m x \sec^n x \, dx$ are

1+\tan^2 x=\sec^2 x,\qquad \frac{d}{dx}\tan x=\sec^2 x,\qquad \frac{d}{dx}\sec x=\tan x\sec x.

[METHOD]

Powers of tangent and secant

To evaluate $\int \tan^m x \sec^n x \, dx$ :

If $n$ is a positive even integer, factor out $\sec^2 x$ and substitute $u=\tan x$ . Convert remaining powers of $\sec$ by $1+\tan^2 x=\sec^2 x$ .
If $n=0$ , use $\tan^2 x=\sec^2 x-1$ to lower the power of $\tan$ by two, with $\int dx$ and $\int\tan x\,dx$ as base cases.
If $n$ $n$ is odd:
- If $m$ is odd, factor out $\tan x\sec x$ and substitute $u=\sec x$ . Convert remaining powers of $\tan$ by $\tan^2 x=\sec^2 x-1$ .
- If $m$ is even, write $\tan^m x=(\sec^2 x-1)^{m/2}$ and expand into powers of $\sec x$ . Reduce odd secant powers by parts until only $\int\sec x\,dx$ remains.

The first rule needs $n$ even: factoring out $\sec^2 x$ when $n$ is odd leaves an unmatched $\sec x$ , so $u=\tan x$ does not absorb the whole integrand.

For the odd secant powers $S_\ell=\int\sec^\ell x\,dx$ , the integration by parts used below for $S_3$ gives the reduction formula

S_\ell=\frac{\sec^{\ell-2}x\tan x}{\ell-1}+\frac{\ell-2}{\ell-1}S_{\ell-2},\qquad \ell\ge3.

To derive it, split $\sec^\ell x=\sec^{\ell-2}x\sec^2x$ to get $S_\ell=\sec^{\ell-2}x\tan x-(\ell-2)\int\sec^{\ell-2}x\tan^2x\,dx$ , substitute $\tan^2x=\sec^2x-1$ and collect the resulting copy of $S_\ell$ . Stop at $S_1=\ln|\sec x+\tan x|$ and add one arbitrary constant at the end.

[1]:Tangent and secant
[2]: $\int \sec^3 x \, dx$
[3]:Average value of a piecewise function
[4]:Seen questions
[5]: $\int \sec^4 x \tan^2 x \, dx$
2 more

[EXAMPLE]

$\int \sec^3 x \, dx$

Write $I=\int \sec^3 x \, dx$ . Split the integrand as $\sec x\cdot\sec^2 x$ and integrate by parts with $u=\sec x$ and $dv=\sec^2 x \, dx$ , so $du=\sec x\tan x \, dx$ and $v=\tan x$ . Then