Trigonometric Integrals

In this section we look at how to integrate a variety of products of trigonometric functions. These integrals are called trigonometric integrals. They are an important part of the integration technique called trigonometric substitution, which is featured in Trigonometric Substitution. This technique allows us to convert algebraic expressions that we may not be able to integrate into expressions involving trigonometric functions, which we may be able to integrate using the techniques described in this section. In addition, these types of integrals appear frequently when we study polar, cylindrical, and spherical coordinate systems later. Let’s begin our study with products of

sin x

Integrating Products and Powers of sinx and cosx

A key idea behind the strategy used to integrate combinations of products and powers of

sin x

involves rewriting these expressions as sums and differences of integrals of the form

\int {sin}^{j} x cos x d x

After rewriting these integrals, we evaluate them using u-substitution. Before describing the general process in detail, let’s take a look at the following examples.

In the next example, we see the strategy that must be applied when there are only even powers of

sin x

are invaluable. These identities are sometimes known as power-reducing identities and they may be derived from the double-angle identity

cos (2 x) = {cos}^{2} x - {sin}^{2} x

Problem-Solving Strategy: Integrating Products and Powers of sin *x* and cos *x*

To integrate $\int {cos}^{j} x {sin}^{k} x d x$

use the following strategies:

If $k$
is odd, rewrite
${sin}^{k} x = {sin}^{k - 1} x sin x$
and use the identity
${sin}^{2} x = 1 - {cos}^{2} x$
to rewrite
${sin}^{k - 1} x$
in terms of
$cos x .$
Integrate using the substitution
$u = cos x .$
This substitution makes
$d u = − sin x d x .$
If $j$
is odd, rewrite
${cos}^{j} x = {cos}^{j - 1} x cos x$
and use the identity
${cos}^{2} x = 1 - {sin}^{2} x$
to rewrite
${cos}^{j - 1} x$
in terms of
$sin x .$
Integrate using the substitution
$u = sin x .$
This substitution makes
$d u = cos x d x .$
(Note: If both
$j$
and
$k$
are odd, either strategy 1 or strategy 2 may be used.)
If both $j$
and
$k$
are even, use
${sin}^{2} x = (1 / 2) - (1 / 2) cos (2 x)$
and
${cos}^{2} x = (1 / 2) + (1 / 2) cos (2 x) .$
After applying these formulas, simplify and reapply strategies 1 through 3 as appropriate.

Integrating

\int {cos}^{j} x {sin}^{k} x d x

where *k* and *j* are Even

Evaluate $\int {sin}^{4} x d x .$

Since the power on $sin x$

is even $(k = 4)$

and the power on $cos x$

is even $(j = 0),$

we must use strategy 3. Thus,

\begin{matrix} \int {sin}^{4} x d x & = \int {({sin}^{2} x)}^{2} d x & Rewrite {sin}^{4} x = {({sin}^{2} x)}^{2} . \\ = \int {(\frac{1}{2} - \frac{1}{2} cos (2 x))}^{2} d x & Substitute {sin}^{2} x = \frac{1}{2} - \frac{1}{2} cos (2 x) . \\ = \int (\frac{1}{4} - \frac{1}{2} cos (2 x) + \frac{1}{4} {cos}^{2} (2 x)) d x & Expand {(\frac{1}{2} - \frac{1}{2} cos (2 x))}^{2} . \\ = \int (\frac{1}{4} - \frac{1}{2} cos (2 x) + \frac{1}{4} (\frac{1}{2} + \frac{1}{2} cos (4 x)) d x . \end{matrix}

Since ${cos}^{2} (2 x)$

has an even power, substitute ${cos}^{2} (2 x) = \frac{1}{2} + \frac{1}{2} cos (4 x) :$

\begin{matrix} = \int (\frac{3}{8} - \frac{1}{2} cos (2 x) + \frac{1}{8} cos (4 x)) d x & Simplify . \\ = \frac{3}{8} x - \frac{1}{4} sin (2 x) + \frac{1}{32} sin (4 x) + C & Evaluate the integral . \end{matrix}

In some areas of physics, such as quantum mechanics, signal processing, and the computation of Fourier series, it is often necessary to integrate products that include

sin (a x),

These integrals are evaluated by applying trigonometric identities, as outlined in the following rule.

These formulas may be derived from the sum-of-angle formulas for sine and cosine.

Integrating Products and Powers of tanx and secx

we rewrite the expression we wish to integrate as the sum or difference of integrals of the form

\int {tan}^{j} x {sec}^{2} x d x

As we see in the following example, we can evaluate these new integrals by using u-substitution.

We now take a look at the various strategies for integrating products and powers of

sec x

Problem-Solving Strategy: Integrating

\int {tan}^{k} x {sec}^{j} x d x

To integrate $\int {tan}^{k} x {sec}^{j} x d x,$

use the following strategies:

If $j$
is even and
$j \geq 2,$
rewrite
${sec}^{j} x = {sec}^{j - 2} x {sec}^{2} x$
and use
${sec}^{2} x = {tan}^{2} x + 1$
to rewrite
${sec}^{j - 2} x$
in terms of
$tan x .$
Let
$u = tan x$
and
$d u = {sec}^{2} x .$
If $k$
is odd and
$j \geq 1,$
rewrite
${tan}^{k} x {sec}^{j} x = {tan}^{k - 1} x {sec}^{j - 1} x sec x tan x$
and use
${tan}^{2} x = {sec}^{2} x - 1$
to rewrite
${tan}^{k - 1} x$
in terms of
$sec x .$
Let
$u = sec x$
and
$d u = sec x tan x d x .$
(Note: If
$j$
is even and
$k$
is odd, then either strategy 1 or strategy 2 may be used.)
If $k$
is odd where
$k \geq 3$
and
$j = 0,$
rewrite
${tan}^{k} x = {tan}^{k - 2} x {tan}^{2} x = {tan}^{k - 2} x ({sec}^{2} x - 1) = {tan}^{k - 2} x {sec}^{2} x - {tan}^{k - 2} x .$
It may be necessary to repeat this process on the
${tan}^{k - 2} x$
term.
If $k$
is even and
$j$
is odd, then use
${tan}^{2} x = {sec}^{2} x - 1$
to express
${tan}^{k} x$
in terms of
$sec x .$
Use integration by parts to integrate odd powers of
$sec x .$

Integrating

\int {sec}^{3} x d x

Integrate $\int {sec}^{3} x d x .$

This integral requires integration by parts. To begin, let $u = sec x$

and $d v = {sec}^{2} x .$

These choices make $d u = sec x tan x$

and $v = tan x .$

Thus,

\begin{matrix} \int {sec}^{3} x d x & = sec x tan x - \int tan x sec x tan x d x \\ = sec x tan x - \int {tan}^{2} x sec x d x & Simplify . \\ = sec x tan x - \int ({sec}^{2} x - 1) sec x d x & Substitute {tan}^{2} x = {sec}^{2} x - 1. \\ = sec x tan x + \int sec x d x - \int {sec}^{3} x d x & Rewrite . \\ = sec x tan x + ln \| sec x + tan x \| - \int {sec}^{3} x d x . & Evaluate \int sec x d x . \end{matrix}

We now have

\int {sec}^{3} x d x = sec x tan x + ln \| sec x + tan x \| - \int {sec}^{3} x d x .

Since the integral $\int {sec}^{3} x d x$

has reappeared on the right-hand side, we can solve for $\int {sec}^{3} x d x$

by adding it to both sides. In doing so, we obtain

2 \int {sec}^{3} x d x = sec x tan x + ln \| sec x + tan x \| .

Dividing by 2, we arrive at

\int {sec}^{3} x d x = \frac{1}{2} sec x tan x + \frac{1}{2} ln \| sec x + tan x \| + C .

Reduction Formulas

is odd requires integration by parts. In addition, we must also know the value of

\int {sec}^{n - 2} x d x

To make the process easier, we can derive and apply the following power reduction formulas. These rules allow us to replace the integral of a power of

sec x

Key Concepts

Key Equations

Fill in the blank to make a true statement.

{sin}^{2} x + \_\_\_\_\_\_\_= 1

{cos}^{2} x

{sec}^{2} x - 1 = \_\_\_\_\_\_\_

Use an identity to reduce the power of the trigonometric function to a trigonometric function raised to the first power.

{sin}^{2} x = \_\_\_\_\_\_\_

\frac{1 - cos (2 x)}{2}

{cos}^{2} x = \_\_\_\_\_\_\_

Evaluate each of the following integrals by u-substitution.

\int {sin}^{3} x cos x d x

\frac{{sin}^{4} x}{4} + C

\int \sqrt{cos x} sin x d x

\int {tan}^{5} (2 x) {sec}^{2} (2 x) d x

\frac{1}{12} {tan}^{6} (2 x) + C

\int {sin}^{7} (2 x) cos (2 x) d x

\int tan (\frac{x}{2}) {sec}^{2} (\frac{x}{2}) d x

{sec}^{2} (\frac{x}{2}) + C

\int {tan}^{2} x {sec}^{2} x d x

Compute the following integrals using the guidelines for integrating powers of trigonometric functions. Use a CAS to check the solutions. (Note: Some of the problems may be done using techniques of integration learned previously.)

\int {sin}^{3} x d x

- \frac{3 cos x}{4} + \frac{1}{12} cos (3 x) + C = − cos x + \frac{{cos}^{3} x}{3} + C

\int {cos}^{3} x d x

\int sin x cos x d x

- \frac{1}{2} {cos}^{2} x + C

\int {cos}^{5} x d x

\int {sin}^{5} x {cos}^{2} x d x

- \frac{5 cos x}{64} - \frac{1}{192} cos (3 x) +

\frac{3}{320} cos (5 x) - \frac{1}{448} cos (7 x) + C

\int {sin}^{3} x {cos}^{3} x d x

\int \sqrt{sin x} cos x d x

\frac{2}{3} {(sin x)}^{2 / 3} + C

\int \sqrt{sin x} {cos}^{3} x d x

\int sec x tan x d x

sec x + C

\int tan (5 x) d x

\int {tan}^{2} x sec x d x

\frac{1}{2} sec x tan x - \frac{1}{2} ln (sec x + tan x) + C

\int tan x {sec}^{3} x d x

\int {sec}^{4} x d x

\frac{2 tan x}{3} + \frac{1}{3} sec {(x)}^{2} tan x

= tan x + \frac{{tan}^{3} x}{3} + C

\int cot x d x

\int csc x d x

− ln \| cot x + csc x \| + C

\int \frac{{tan}^{3} x}{\sqrt{sec x}} d x

For the following exercises, find a general formula for the integrals.

\int {sin}^{2} a x cos a x d x

\frac{{sin}^{3} (a x)}{3 a} + C

\int sin a x cos a x d x .

Use the double-angle formulas to evaluate the following integrals.

\int_{0}^{π} {sin}^{2} x d x

\frac{π}{2}

\int_{0}^{π} {sin}^{4} x d x

\int {cos}^{2} 3 x d x

\frac{x}{2} + \frac{1}{12} sin (6 x) + C

\int {sin}^{2} x {cos}^{2} x d x

\int {sin}^{2} x d x + \int {cos}^{2} x d x

$x + C$

\int {sin}^{2} x {cos}^{2} (2 x) d x

For the following exercises, evaluate the definite integrals. Express answers in exact form whenever possible.

\int_{0}^{2 π} cos x sin 2 x d x

\int_{0}^{π} sin 3 x sin 5 x d x

\int_{0}^{π} cos (99 x) sin (101 x) d x

\int_{− π}^{π} {cos}^{2} (3 x) d x

\int_{0}^{2 π} sin x sin (2 x) sin (3 x) d x

\int_{0}^{4 π} cos (x / 2) sin (x / 2) d x

\int_{π / 6}^{π / 3} \frac{{cos}^{3} x}{\sqrt{sin x}} d x

(Round this answer to three decimal places.)

Approximately 0.239

\int_{− π / 3}^{π / 3} \sqrt{{sec}^{2} x - 1} d x

\int_{0}^{π / 2} \sqrt{1 - cos (2 x)} d x

\sqrt{2}

Find the area of the region bounded by the graphs of the equations $y = sin x, y = {sin}^{3} x, x = 0, and x = \frac{π}{2} .$

Find the area of the region bounded by the graphs of the equations $y = {cos}^{2} x, y = {sin}^{2} x, x = - \frac{π}{4}, and x = \frac{π}{4} .$

1.0

A particle moves in a straight line with the velocity function $v (t) = sin (ω t) {cos}^{2} (ω t) .$

Find its position function $x = f (t)$

if $f (0) = 0 .$

Find the average value of the function $f (x) = {sin}^{2} x {cos}^{3} x$

over the interval $[− π, π] .$

For the following exercises, solve the differential equations.

\frac{d y}{d x} = {sin}^{2} x .

The curve passes through point $(0, 0) .$

\frac{d y}{d θ} = {sin}^{4} (π θ)

\frac{3 θ}{8} - \frac{1}{4 π} sin (2 π θ) + \frac{1}{32 π} sin (4 π θ) + C = f (x)

Find the length of the curve $y = ln (csc x), \frac{π}{4} \leq x \leq \frac{π}{2} .$

Find the length of the curve $y = ln (sin x), \frac{π}{3} \leq x \leq \frac{π}{2} .$

ln (\sqrt{3})

Find the volume generated by revolving the curve $y = cos (3 x)$

about the x-axis, $0 \leq x \leq \frac{π}{36} .$

For the following exercises, use this information: The inner product of two functions f and g over $[a, b]$

is defined by $f (x) \cdot g (x) = 〈 f, g 〉 = \int_{a}^{b} f \cdot g d x .$

Two distinct functions f and g are said to be orthogonal if $〈 f, g 〉 = 0 .$

Show that ${sin (2 x), cos (3 x)}$

are orthogonal over the interval $[− π, π] .$

\int_{− π}^{π} sin (2 x) cos (3 x) d x = 0

Evaluate $\int_{− π}^{π} sin (m x) cos (n x) d x .$

Integrate $y^{'} = \sqrt{tan x} {sec}^{4} x .$

\sqrt{tan (x)} x (\frac{8 tan x}{21} + \frac{2}{7} sec x^{2} tan x) + C = f (x)

For each pair of integrals, determine which one is more difficult to evaluate. Explain your reasoning.

\int {sin}^{456} x cos x d x

or $\int {sin}^{2} x {cos}^{2} x d x$

\int {tan}^{350} x {sec}^{2} x d x

or $\int {tan}^{350} x sec x d x$

The second integral is more difficult because the first integral is simply a u-substitution type.