Integrating Products and Powers of tanx and secx

Learning Outcomes

Solve integration problems involving products and powers of $\tan{x}$ and $\sec{x}$
Use reduction formulas to solve trigonometric integrals

Before discussing the integration of products and powers of $\tan{x}$ and $\sec{x}$ , it is useful to recall the integrals involving $\tan{x}$ and $\sec{x}$ we have already learned:

${\displaystyle\int}{\sec}^{2}xdx=\tan{x}+C$
${\displaystyle\int}\sec{x}\tan{x}dx=\sec{x}+C$
${\displaystyle\int}\tan{x}dx=\text{ln}|\sec{x}|+C$
${\displaystyle\int}\sec{x}dx=\text{ln}|\sec{x}+\tan{x}|+C$ .

For most integrals of products and powers of $\tan{x}$ and $\sec{x}$ , we rewrite the expression we wish to integrate as the sum or difference of integrals of the form ${\displaystyle\int}{\tan}^{j}x{\sec}^{2}xdx$ or ${\displaystyle\int}{\sec}^{j}x\tan{x}dx$ . As we see in the following example, we can evaluate these new integrals by using u-substitution. Before doing so, it is useful to note how the Pythagorean Identity implies relationships between other pairs of trigonometric functions.

Recall: The Pythagorean Identity

For any angle $x$ :

$\sin^2 x + \cos^2 x = 1$

Dividing the original equation by $\cos^2 x$ and simplifying yields an expression for $\sec^2 x$ in terms of $\tan^2 x$ :

$\tan^2 x + 1 = \sec^2 x$

Subtracting both sides of the equation by $1$ yields an expression for $\tan^2 x$ in terms of $\sec^2 x$ :

$\tan^2 x = \sec^2 x - 1$

Example: Evaluating ${\displaystyle\int}{\sec}^{j}x\tan{x}dx$

Evaluate ${\displaystyle\int}{\sec}^{5}x\tan{x}dx$ .

Show Solution

Start by rewriting ${\sec}^{5}x\tan{x}$ as ${\sec}^{4}x\sec{x}\tan{x}$ .

\begin{matrix} \int {sec}^{5} x tan x d x & = \int {sec}^{4} x sec x tan x d x & Let u = sec x; then, d u = sec x tan x d x . = \int u^{4} d u & Evaluate the integral . = \frac{1}{5} u^{5} + C & Substitute sec x = u . = \frac{1}{5} {sec}^{5} x + C \end{matrix}

$\begin{array}{cccc}\hfill {\displaystyle\int}{\sec}^{5}x\tan{x}dx& ={\displaystyle\int}{\sec}^{4}x\sec{x}\tan{x}dx\hfill & & \text{Let }u=\sec{x};\text{then},du=\sec{x}\tan{x}dx.\hfill \\ & ={\displaystyle\int}{u}^{4}du\hfill & & \text{Evaluate the integral}.\hfill \\ & =\frac{1}{5}{u}^{5}+C\hfill & & \text{Substitute }\sec{x}=u.\hfill \\ & =\frac{1}{5}{\sec}^{5}x+C\hfill & & \end{array}$

Interactive

You can read some interesting information at this website to learn about a common integral involving the secant.

try it

Evaluate ${\displaystyle\int}{\tan}^{5}x{\sec}^{2}xdx$ .

Hint

Let $u=\tan{x}$ and $du={\sec}^{2}x$ .

Show Solution

$\frac{1}{6}{\tan}^{6}x+C$

We now take a look at the various strategies for integrating products and powers of $\sec{x}$ and $\tan{x}$ .

Problem-Solving Strategy: Integrating ${\displaystyle\int}{\tan}^{k}x{\sec}^{j}xdx$

To integrate ${\displaystyle\int}{\tan}^{k}x{\sec}^{j}xdx$ , use the following strategies:

If $j$ is even and $j\ge 2$ , 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 $du={\sec}^{2}x$ .
If $k$ is odd and $j\ge 1$ , 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 $du=\sec{x}\tan{x}dx$ . (Note: If $j$ is even and $k$ is odd, then either strategy 1 or strategy 2 may be used.)
If $k$ is odd where $k\ge 3$ and $j=0$ , rewrite ${\tan}^{k}x={\tan}^{k - 2}x{\tan}^{2}x={\tan}^{k - 2}x\left({\sec}^{2}x - 1\right)={\tan}^{k - 2}x{\sec}^{2}x-{\tan}^{k - 2}x$ . 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 $\sec{x}$ .

Example: Integrating ${\displaystyle\int}{\tan}^{k}x{\sec}^{j}xdx$ when $j$ is Even

Evaluate ${\displaystyle\int}{\tan}^{6}x{\sec}^{4}xdx$ .

Show Solution

Since the power on $\sec{x}$ is even, rewrite ${\sec}^{4}x={\sec}^{2}x{\sec}^{2}x$ and use ${\sec}^{2}x={\tan}^{2}x+1$ to rewrite the first ${\sec}^{2}x$ in terms of $\tan{x}$ . Thus,

\begin{matrix} \int {tan}^{6} x {sec}^{4} x d x & = \int {tan}^{6} x ({tan}^{2} x + 1) {sec}^{2} x d x & Let u = tan x and d u = {sec}^{2} x . = \int u^{6} (u^{2} + 1) d u & Expand . = \int (u^{8} + u^{6}) d u & Evaluate the integral . = \frac{1}{9} u^{9} + \frac{1}{7} u^{7} + C & Substitute tan x = u . = \frac{1}{9} {tan}^{9} x + \frac{1}{7} {tan}^{7} x + C . \end{matrix}

$\begin{array}{cccc}\hfill {\displaystyle\int}{\tan}^{6}x{\sec}^{4}xdx& ={\displaystyle\int}{\tan}^{6}x\left({\tan}^{2}x+1\right){\sec}^{2}xdx\hfill & & \text{Let }u=\tan{x}\text{ and }du={\sec}^{2}x.\hfill \\ & ={\displaystyle\int}{u}^{6}\left({u}^{2}+1\right)du\hfill & & \text{Expand}.\hfill \\ & ={\displaystyle\int}\left({u}^{8}+{u}^{6}\right)du\hfill & & \text{Evaluate the integral}.\hfill \\ & =\frac{1}{9}{u}^{9}+\frac{1}{7}{u}^{7}+C\hfill & & \text{Substitute }\tan{x}=u.\hfill \\ & =\frac{1}{9}{\tan}^{9}x+\frac{1}{7}{\tan}^{7}x+C.\hfill & & \hfill \end{array}$

Example: Integrating ${\displaystyle\int}{\tan}^{k}x{\sec}^{j}xdx$ when $k$ is Odd

Evaluate ${\displaystyle\int}{\tan}^{5}x{\sec}^{3}xdx$ .

Show Solution

Since the power on $\tan{x}$ is odd, begin by rewriting ${\tan}^{5}x{\sec}^{3}x={\tan}^{4}x{\sec}^{2}x\sec{x}\tan{x}$ . Thus,

\begin{matrix} {tan}^{5} x {sec}^{3} x & = & {tan}^{4} x {sec}^{2} x sec x tan x . & Write {tan}^{4} x = {({tan}^{2} x)}^{2} . \int {tan}^{5} x {sec}^{3} x d x & = & \int {({tan}^{2} x)}^{2} {sec}^{2} x sec x tan x d x & Use {tan}^{2} x = {sec}^{2} x - 1. = & \int {({sec}^{2} x - 1)}^{2} {sec}^{2} x sec x tan x d x & Let u = sec x and d u = sec x tan x d x . = & \int {(u^{2} - 1)}^{2} u^{2} d u & Expand . = & \int (u^{6} - 2 u^{4} + u^{2}) d u & Integrate . = & \frac{1}{7} u^{7} - \frac{2}{5} u^{5} + \frac{1}{3} u^{3} + C & Substitute sec x = u . = & \frac{1}{7} {sec}^{7} x - \frac{2}{5} {sec}^{5} x + \frac{1}{3} {sec}^{3} x + C . \end{matrix}

$\begin{array}{cccccc}\hfill {\tan}^{5}x{\sec}^{3}x& =\hfill & {\tan}^{4}x{\sec}^{2}x\sec{x}\tan{x}.\hfill & & & \text{Write }{\tan}^{4}x={\left({\tan}^{2}x\right)}^{2}.\hfill \\ \hfill {\displaystyle\int}{\tan}^{5}x{\sec}^{3}xdx& =\hfill & {\displaystyle\int}{\left({\tan}^{2}x\right)}^{2}{\sec}^{2}x\sec{x}\tan{x}dx\hfill & & & \text{Use }{\tan}^{2}x={\sec}^{2}x - 1.\hfill \\ & =\hfill & {\displaystyle\int}{\left({\sec}^{2}x - 1\right)}^{2}{\sec}^{2}x\sec{x}\tan{x}dx\hfill & & & \text{Let }u=\sec{x}\text{and}du=\sec{x}\tan{x}dx.\hfill \\ & =\hfill & {\displaystyle\int}{\left({u}^{2}-1\right)}^{2}{u}^{2}du\hfill & & & \text{Expand}.\hfill \\ & =\hfill & {\displaystyle\int}\left({u}^{6}-2{u}^{4}+{u}^{2}\right)du\hfill & & & \text{Integrate}.\hfill \\ & =\hfill & \frac{1}{7}{u}^{7}-\frac{2}{5}{u}^{5}+\frac{1}{3}{u}^{3}+C\hfill & & & \text{Substitute }\sec{x}=u.\hfill \\ & =\hfill & \frac{1}{7}{\sec}^{7}x-\frac{2}{5}{\sec}^{5}x+\frac{1}{3}{\sec}^{3}x+C.\hfill & & & \end{array}$

Example: Integrating ${\displaystyle\int}{\tan}^{k}xdx$ where $k$ is Odd and $k\ge 3$

Evaluate ${\displaystyle\int}{\tan}^{3}xdx$ .

Show Solution

Begin by rewriting ${\tan}^{3}x=\tan{x}{\tan}^{2}x=\tan{x}\left({\sec}^{2}x - 1\right)=\tan{x}{\sec}^{2}x-\tan{x}$ . Thus,

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

$\begin{array}{cc}{\displaystyle\int}{\tan}^{3}xdx\hfill & ={\displaystyle\int}\left(\tan{x}{\sec}^{2}x-\tan{x}\right)dx\hfill \\ \hfill & ={\displaystyle\int}\tan{x}{\sec}^{2}xdx-{\displaystyle\int}\tan{x}dx\hfill \\ \hfill & =\frac{1}{2}{\tan}^{2}x-\text{ln}|\sec{x}|+C.\hfill \end{array}$

For the first integral, use the substitution $u=\tan{x}$ . For the second integral, use the formula.

Example: Integrating ${\displaystyle\int}{\sec}^{3}xdx$

Integrate ${\displaystyle\int}{\sec}^{3}xdx$ .

Show Solution

This integral requires integration by parts. To begin, let $u=\sec{x}$ and $dv={\sec}^{2}x$ . These choices make $du=\sec{x}\tan{x}$ 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}

$\begin{array}{cccc}\hfill {\displaystyle\int}{\sec}^{3}xdx& =\sec{x}\tan{x}-{\displaystyle\int}\tan{x}\sec{x}\tan{x}dx\hfill & & \\ & =\sec{x}\tan{x}-{\displaystyle\int}{\tan}^{2}x\sec{x}dx\hfill & & \text{Simplify}.\hfill \\ & =\sec{x}\tan{x}-{\displaystyle\int}\left({\sec}^{2}x - 1\right)\sec{x}dx\hfill & & \text{Substitute }{\tan}^{2}x={\sec}^{2}x - 1.\hfill \\ & =\sec{x}\tan{x}+{\displaystyle\int}\sec{x}dx-{\displaystyle\int}{\sec}^{3}xdx\hfill & & \text{Rewrite}.\hfill \\ & =\sec{x}\tan{x}+\text{ln}|\sec{x}+\tan{x}|-{\displaystyle\int}{\sec}^{3}xdx.\hfill & & \text{Evaluate}{\displaystyle\int}\sec{x}dx.\hfill \end{array}$

We now have

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

${\displaystyle\int}{\sec}^{3}xdx=\sec{x}\tan{x}+\text{ln}|\sec{x}+\tan{x}|-{\displaystyle\int}{\sec}^{3}xdx$ .

Since the integral ${\displaystyle\int}{\sec}^{3}xdx$ has reappeared on the right-hand side, we can solve for ${\displaystyle\int}{\sec}^{3}xdx$ 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 |

$2{\displaystyle\int}{\sec}^{3}xdx=\sec{x}\tan{x}+\text{ln}|\sec{x}+\tan{x}|$ .

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

${\displaystyle\int}{\sec}^{3}xdx=\frac{1}{2}\sec{x}\tan{x}+\frac{1}{2}\text{ln}|\sec{x}+\tan{x}|+C$ .

try it

Evaluate ${\displaystyle\int}{\tan}^{3}x{\sec}^{7}xdx$ .

Hint

Use the previous example as a guide: Integrating ${\displaystyle\int}{\tan}^{k}x{\sec}^{j}xdx$ when $k$ is Odd.

Show Solution

$\frac{1}{9}{\sec}^{9}x-\frac{1}{7}{\sec}^{7}x+C$

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Reduction Formulas

Evaluating ${\displaystyle\int}{\sec}^{n}xdx$ for values of $n$ where $n$ is odd requires integration by parts. In addition, we must also know the value of ${\displaystyle\int}{\sec}^{n - 2}xdx$ to evaluate ${\displaystyle\int}{\sec}^{n}xdx$ . The evaluation of ${\displaystyle\int}{\tan}^{n}xdx$ also requires being able to integrate ${\displaystyle\int}{\tan}^{n - 2}xdx$ . To make the process easier, we can derive and apply the following . These rules allow us to replace the integral of a power of $\sec{x}$ or $\tan{x}$ with the integral of a lower power of $\sec{x}$ or $\tan{x}$ .

Rule: Reduction Formulas for ${\displaystyle\int}{\sec}^{n}xdx$ and ${\displaystyle\int}{\tan}^{n}xdx$

\int {sec}^{n} x d x = \frac{1}{n - 1} {sec}^{n - 2} x tan x + \frac{n - 2}{n - 1} \int {sec}^{n - 2} x d x

${\displaystyle\int}{\sec}^{n}xdx=\frac{1}{n - 1}{\sec}^{n - 2}x\tan{x}+\frac{n - 2}{n - 1}{\displaystyle\int}{\sec}^{n - 2}xdx$

\int {tan}^{n} x d x = \frac{1}{n - 1} {tan}^{n - 1} x - \int {tan}^{n - 2} x d x

${\displaystyle\int}{\tan}^{n}xdx=\frac{1}{n - 1}{\tan}^{n - 1}x-{\displaystyle\int}{\tan}^{n - 2}xdx$

The first power reduction rule may be verified by applying integration by parts. The second may be verified by following the strategy outlined for integrating odd powers of $\tan{x}$ .

Example: Revisiting ${\displaystyle\int}{\sec}^{3}xdx$

Apply a reduction formula to evaluate ${\displaystyle\int}{\sec}^{3}xdx$ .

Show Solution

By applying the first reduction formula, we obtain

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

$\begin{array}{cc}{\displaystyle\int}{\sec}^{3}xdx\hfill & =\frac{1}{2}\sec{x}\tan{x}+\frac{1}{2}{\displaystyle\int}\sec{x}dx\hfill \\ \hfill & =\frac{1}{2}\sec{x}\tan{x}+\frac{1}{2}\text{ln}|\sec{x}+\tan{x}|+C.\hfill \end{array}$

Example: Using a Reduction Formula

Evaluate ${\displaystyle\int}{\tan}^{4}xdx$ .

Show Solution

Applying the reduction formula for ${\displaystyle\int}{\tan}^{4}xdx$ we have

\begin{matrix} \int {tan}^{4} x d x & = \frac{1}{3} {tan}^{3} x - \int {tan}^{2} x d x = \frac{1}{3} {tan}^{3} x - (tan x - \int {tan}^{0} x d x) & Apply the reduction formula to \int {tan}^{2} x d x . = \frac{1}{3} {tan}^{3} x - tan x + \int 1 d x & Simplify . = \frac{1}{3} {tan}^{3} x - tan x + x + C . & Evaluate \int 1 d x . \end{matrix}

$\begin{array}{cccc}\hfill {\displaystyle\int}{\tan}^{4}xdx& =\frac{1}{3}{\tan}^{3}x-{\displaystyle\int}{\tan}^{2}xdx\hfill & & \\ & =\frac{1}{3}{\tan}^{3}x-\left(\tan{x}-{\displaystyle\int}{\tan}^{0}xdx\right)\hfill & & \text{Apply the reduction formula to}{\displaystyle\int}{\tan}^{2}xdx.\hfill \\ & =\frac{1}{3}{\tan}^{3}x-\tan{x}+{\displaystyle\int}1dx\hfill & & \text{Simplify}.\hfill \\ & =\frac{1}{3}{\tan}^{3}x-\tan{x}+x+C.\hfill & & \text{Evaluate}{\displaystyle\int}1dx.\hfill \end{array}$

try it

Apply the reduction formula to ${\displaystyle\int}{\sec}^{5}xdx$ .

Hint

Use reduction formula 1 and let $n=5$ .

Show Solution

${\displaystyle\int}{\sec}^{5}xdx=\frac{1}{4}{\sec}^{3}x\tan{x}-\frac{3}{4}{\displaystyle\int}{\sec}^{3}x$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for this segmented clip of “3.2 Trigonometric Integrals” here (opens in new window).

Module 3: Techniques of Integration