(AI3) Trigonometric Integrals (Powers of Secant and Tangent)#

In this lesson we are going to see how to calculate integrals of the form:

\int \tan^{m} x \sec^{n} x d x

Videos#

Part 1 of 2

Part 2 of 2

Substitution Rule#

Our main strategy is going to be a $u$ -substitution with either $u = \sec x$ or $u = \tan x$ . Which one we choose, ultimately comes down to the differentials.

secant

\begin{aligned} u & = \sec x \\ d u & = \sec x \tan x d x \end{aligned}

tangent

\begin{aligned} u & = \tan x \\ d u & = \sec^{2} x d x \end{aligned}

In order to make the differentials match with what’s actually in our integral, we start with the function we’re integrating and factor off either a $(\sec x \tan x)$ term or a $(\sec^{2} x)$ term and put it next to the $d x$ .

In order to convert between powers of $\sec x$ and powers of $\tan x$ we use the following identity:

Trigonometric Identity

\sec^{2} θ = 1 + \tan^{2} θ

Example 1#

Integrate the following:

\int \tan^{6} x \sec^{4} x d x

Solution (Click to see the steps.)

Step 1

Separate: We start by separating off a $(\sec^{2} x)$ term to get:

\int \tan^{6} x \sec^{4} x d x = \int \tan^{6} x \sec^{2} x \cdot (\sec^{2} x) d x

Step 2

Convert: Since we factored off a $(\sec^{2} x)$ term, this means we need to convert all remaining trig functions into powers of $\tan x$ .

\begin{aligned} = \int \tan^{6} x \sec^{2} x \cdot (\sec^{2} x) d x \\ = \int \tan^{6} x (1 + \tan^{2} x) \cdot (\sec^{2} x) d x \end{aligned}

This is where we use the trig identity: $\sec^{2} θ = 1 + \tan^{2} θ$

Step 3

Choose u: Now we’re ready for a $u$ -substitution. Since we factored off a $(\sec^{2} x)$ term, we choose:

u = \tan x

and then we calculate $d u$ to get:

d u = \sec^{2} x d x

And then substituting our $u$ and $d u$ into the integral gives us:

\begin{aligned} = \int \tan^{6} x (1 + \tan^{2} x) \cdot (\sec^{2} x) d x \\ = \int u^{6} (1 + u^{2}) d u \end{aligned}

Step 4

Finish It! And finally we finish the integration. First multiply everything out so that we have only power functions to integrate.

\begin{aligned} = \int u^{6} (1 + u^{2}) d u \\ = \int (u^{6} + u^{8}) d u \\ = \frac{1}{7} u^{7} + \frac{1}{9} u^{9} + C \end{aligned}

And then convert back to $x$ to get our answer:

\begin{aligned} = \frac{1}{7} \tan^{7} x + \frac{1}{9} \tan^{9} x + C \end{aligned}

(Don’t forget $+ C$ for indefinite integrals!)

Example 2#

Integrate the following:

\int \tan^{5} x \sec^{7} x d x

Solution (Click to see the steps.)

Step 1

Separate: We start by separating off a $(\sec x \tan x)$ term to get:

\int \tan^{5} x \sec^{7} x d x = \int \tan^{4} x \sec^{6} x \cdot (\sec x \tan x) d x

Step 2

Convert: Since we factored off a $(\sec x \tan x)$ term, this means we need to convert all remaining trig functions into powers of $\sec x$ .

\begin{aligned} = \int \tan^{4} x \cdot \sec^{6} x \cdot (\sec x \tan x) d x \\ = \int \tan^{2} x \cdot \tan^{2} x \cdot \sec^{6} x \cdot (\sec x \tan x) d x \\ = \int (\sec^{2} x - 1) (\sec^{2} x - 1) \sec^{6} x \cdot (\sec x \tan x) d x \end{aligned}

This is where we use the trig identity: $\sec^{2} θ = 1 + \tan^{2} θ$

Step 3

Choose u: Now we’re ready for a $u$ -substitution. Since we factored off a $(\sec x \tan x)$ term, we choose:

u = \sec x

and then we calculate $d u$ to get:

d u = \sec x \tan x d x

And then substituting our $u$ and $d u$ into the integral gives us:

\begin{aligned} = \int (\sec^{2} x - 1) (\sec^{2} x - 1) \sec^{6} x \cdot (\sec x \tan x) d x \\ = \int (u^{2} - 1) (u^{2} - 1) u^{6} d u \end{aligned}

Step 4

Finish It! And finally we finish the integration. First multiply everything out so that we have power functions to integrate.

\begin{aligned} = \int (u^{2} - 1) (u^{2} - 1) u^{6} d u \\ = \int (u^{4} - 2 u^{2} + 1) u^{6} d u \\ = \int (u^{10} - 2 u^{8} + u^{6}) d u \\ = \frac{1}{11} u^{11} - \frac{2}{9} u^{9} + \frac{1}{7} u^{7} + C \end{aligned}

And then finally we convert back to $x$ to get our answer:

\begin{aligned} = \frac{1}{11} \sec^{11} x - \frac{2}{9} \sec^{9} x + \frac{1}{7} \sec^{7} x + C \end{aligned}

(Don’t forget $+ C$ for indefinite integrals!)

Get Creative#

Unfortunately, the substitution strategy used in the first two examples doesn’t always work. If this happens we have a few more things to try:

Use Trig Identities: Keep trying to use our main identity $\sec^{2} θ = 1 + \tan^{2} θ$ , you might still get something useful.
Convert to sine and cosine: We have techniques for integrating powers of $\sin x$ and $\cos x$ which might be helpful.
Integration by parts: This can be a helpful way to move exponents between trig functions. Maybe try making the $d v$ term either $\sec^{2} x$ or $\sec x \tan x$ .

A lot of times, success with the above methods comes down to also knowing:

Helpful Antiderivatives

\begin{aligned} \int \tan x d x & = \ln | \sec x | + C \\ \int \sec x d x & = \ln | \sec x + \tan x | + C \end{aligned}

Example 3#

Integrate the following:

\int \tan^{3} x d x

Solution 1 (Trig identities)

Step 1

Separate: We are still going to try using our trig identity: $\sec^{2} x = 1 + \tan^{2} x$ . Since this has a $\tan^{2} x$ term, let’s start by breaking our function up:

\int \tan^{3} x d x = \int \tan x \tan^{2} x d x

Step 2

Trig Identity: Now we can apply the trig identity $\sec^{2} x = 1 + \tan^{2} x$ to get:

\begin{aligned} \int \tan x \tan^{2} x d x & = \int \tan x (\sec^{2} x - 1) d x \\ = \int (\tan x \cdot \sec^{2} x - \tan x) d x \end{aligned}

And we also multiply everything out.

Step 3

Now it comes down to 2 integrals:

Step 4

Finish It! And finally we put it all together and get:

\begin{aligned} = \int (\tan x \cdot \sec^{2} x - \tan x) d x \\ = \int \tan x \cdot \sec^{2} x d x - \int \tan x d x \\ = \frac{1}{2} \tan^{2} x - \ln | \sec x | + C \end{aligned}

(Don’t forget $+ C$ for indefinite integrals!)

Solution 2 (Convert to sine and cosine)

Step 1

Convert: We are going to start by converting everything into sine and cosine terms:

\int \tan^{3} x d x = \int \frac{\sin^{3} x}{\cos^{3} x} d x

Step 2

Powers of Sine: Since we have an odd power of sine in the numerator, we can separate off one $\sin x$ and put it next to the $d x$

\int \frac{\sin^{3} x}{\cos^{3} x} d x = \int \frac{\sin^{2} x}{\cos^{3} x} \cdot \sin x d x

Putting a $\sin x$ next to $d x$ tells us we are going to use the opposite function $\cos x$ as our $u$ -sub. So let’s convert everything else remaining in our integral to cosine:

= \int \frac{1 - \cos^{2} x}{\cos^{3} x} \cdot \sin x d x

Here we use our standard trig identity: $\sin^{2} θ + \cos^{2} θ = 1$

Step 3

Substitution: Now we can do our $u$ -substitution:

\begin{aligned} u & = \cos x \\ d u & = - \sin x d x ⟶ (- 1) d u = \sin x d x \end{aligned}

And plugging all of this into our integral gives us:

\begin{aligned} = \int \frac{1 - \cos^{2} x}{\cos^{3} x} \cdot \sin x d x \\ = \int \frac{1 - u^{2}}{u^{3}} \cdot (- 1) d u \\ = \int \frac{u^{2} - 1}{u^{3}} d u \end{aligned}

Step 4

Finish It! And finally if we do the division we can finish the integration:

\begin{aligned} = \int \frac{u^{2} - 1}{u^{3}} d u \\ = \int \frac{1}{u} - \frac{1}{u^{3}} d u \\ = \ln | u | + \frac{1}{2} u^{- 2} + C \end{aligned}

And converting back to our original variable:

\begin{aligned} = \ln | \cos x | + \frac{1}{2} \cos^{- 2} x + C \end{aligned}

(Don’t forget $+ C$ for indefinite integrals!)

Example 4#

Integrate the following:

\int \sec^{3} x d x

Solution (Using Integration By Parts and Trig Identities)

Step 1

Separate: We start by separating off a $(\sec^{2} x)$ term to get:

\int \sec^{3} x d x = \int \sec x \cdot \sec^{2} x d x

Now we can choose $u$ and $d v$ :

\begin{aligned} u & = \sec x \\ d u & = \sec x \tan x d x \end{aligned}

\begin{aligned} d v & = \sec^{2} x d x \\ v & = \tan x \end{aligned}

The reason why we factored off a $\sec^{2} x$ term and chose that for $d v$ is because it is easy to integrate (its one of our elementary antiderivatives).

Applying the integration by parts formula gives us:

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

Step 2

Trig Identity: The resulting integral from the integration by parts is still tricky, so we try using a trig identity:

\begin{aligned} = \sec x \tan x - \int \sec x \tan^{2} x d x \\ = \sec x \tan x - \int \sec x (\sec^{2} x - 1) d x \\ = \sec x \tan x - \int \sec^{3} x - \sec x d x \end{aligned}

And now it seems like we might be back where we started, so let’s formally state what we have so far:

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

Step 3

Unknown Integral: If we look at this resulting equation we see that our unknown integral shows up on both sides.

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

So let’s actually solve for it! And to help picture some of the algebra we’ll rename our unknown integral:

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

Giving us the (shortened) equation:

Y = \sec x \tan x - Y + \int \sec x d x

Now we just need to solve for $Y$ . Add $Y$ to both sides to get:

2 Y = \sec x \tan x + \int \sec x d x

And then divide both sides by $2$ :

Y = \frac{1}{2} \sec x \tan x + \frac{1}{2} \int \sec x d x

And there we go!

\int \sec^{3} x d x = \frac{1}{2} \sec x \tan x + \frac{1}{2} \int \sec x d x

Step 4

Finish It! And finally we can finish the remaining integral with the use of some helpful antiderivatives:

\begin{aligned} \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{aligned}

(Don’t forget $+ C$ for indefinite integrals!)

Helpful Antiderivatives

\begin{aligned} \int \tan x d x & = \ln | \sec x | + C \\ \int \sec x d x & = \ln | \sec x + \tan x | + C \end{aligned}

(AI3) Trigonometric Integrals (Powers of Secant and Tangent)

Contents

(AI3) Trigonometric Integrals (Powers of Secant and Tangent)#

Videos#

Substitution Rule#

Example 1#

Example 2#

Get Creative#

Example 3#

Example 4#