2.2: Trigonometric Integrals

Learning Objectives

• Solve integration problems involving products and powers of \(\sin x\) and \(\cos x\).
• Solve integration problems involving products and powers of \(\tan x\) and \(\sec x\).
• Use reduction formulas to solve trigonometric integrals.

In this section we look at how to integrate a variety of products of trigonometric functions. As a collection, these integrals are called trigonometric integrals. They are an important part of the integration technique called trigonometric substitution, which is featured in Section 2.3: 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 \(\sin x\) and \(\cos x.\)

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Integrating Products and Powers of \(\sin(x)\) and \(\cos(x)\)

A key idea behind the strategy used to integrate combinations of products and powers of \(\sin(x)\) and \(\cos(x)\) involves rewriting these expressions as sums and differences of integrals of the form \( \displaystyle \int \sin^j(x) \cos(x)\,dx\) or \( \displaystyle \int \cos^j(x) \sin(x)\,dx\). 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.

Example \(\PageIndex{1}\): Integrating \(\cos^j(x)\sin(x)\)

Evaluate \(\displaystyle \int \cos^3(x)\sin(x)\,dx.\)

Solution

Use \(u\)-substitution and let \(u=\cos(x)\). In this case, \(du=−\sin(x)\,dx.\)

Thus,

\[ \int \cos^3(x)\sin(x)\,dx=− \int u^3\,du=−\dfrac{1}{4}u^4+C=−\dfrac{1}{4}\cos^4(x)+C.\nonumber \]

Exercise \(\PageIndex{1}\)

Evaluate \(\displaystyle \int \sin^4(x)\cos(x)\,dx.\)

Hint

Let \(u=\sin(x).\)

Answer

\(\displaystyle \int \sin^4(x) \cos(x)\,dx = \frac{1}{5}\sin^5x+C\)

Example \(\PageIndex{2}\): A Preliminary Example: Integrating \( \cos^j(x)\sin^k(x)\) where \(k\) is Odd

Evaluate \(\displaystyle \int \cos^2(x)\sin^3(x)\,dx.\)

Solution

To convert this integral to integrals of the form \(\displaystyle \int \cos^j(x)\sin(x)\,dx\), rewrite \(\sin^3(x)=\sin^2(x)\sin(x)\) and make the substitution \(\sin^2(x)=1−\cos^2(x).\)

Thus,

\[\begin{array}{rcl}
\displaystyle \int \cos^2(x)\sin^3(x)\,dx & = & \displaystyle \int \cos^2(x)(1−\cos^2(x))\sin(x)\,dx \\
& = & − \displaystyle \int u^2(1−u^2)\,du \\
& = & \displaystyle \int (u^4−u^2)\,du \\
& = & \dfrac{1}{5}u^5−\dfrac{1}{3}u^3+C \\
& = & \dfrac{1}{5}\cos^5(x)−\dfrac{1}{3}\cos^3(x)+C. \\
\end{array} \nonumber \]

Exercise \(\PageIndex{2}\)

Evaluate \(\displaystyle \int \cos^3(x)\sin^2(x)\,dx.\)

Hint

Write \(\cos^3(x)=\cos^2(x)\cos(x)=(1−\sin^2(x))\cos(x)\) and let \(u=\sin(x)\).

Answer

\(\displaystyle \int \cos^3(x)\sin^2(x)\,dx = \frac{1}{3}\sin^3(x)−\frac{1}{5}\sin^5(x)+C\)

In the next example, we see the strategy that must be applied when there are only even powers of \(\sin(x)\) and \(\cos(x)\). For integrals of this type, the identities

\[\sin^2(x)=\dfrac{1}{2}−\dfrac{1}{2}\cos(2x)=\dfrac{1−\cos(2x)}{2} \nonumber \]

and

\[\cos^2(x)=\dfrac{1}{2}+\dfrac{1}{2}\cos(2x)=\dfrac{1+\cos(2x)}{2} \nonumber \]

are invaluable. These identities are sometimes known as the Power Reduction Formulas and they may be derived from the Double-Angle Identity \(\cos(2x)=\cos^2(x)−\sin^2(x)\) and the Pythagorean Identity \(\cos^2(x)+\sin^2(x)=1.\)

Example \(\PageIndex{3}\): Integrating an Even Power of \(\sin(x)\)

Evaluate \(\displaystyle \int \sin^2(x)\,dx\).

Solution

To evaluate this integral, let’s use the trigonometric identity \(\sin^2(x)=\frac{1}{2}−\frac{1}{2}\cos(2x).\) Thus,

\[ \int \sin^2(x)\,dx= \int \left(\frac{1}{2}−\frac{1}{2}\cos(2x)\right)\,dx=\frac{1}{2}x−\frac{1}{4}\sin(2x)+C.\nonumber \]

Exercise \(\PageIndex{3}\)

Evaluate \(\displaystyle \int \cos^2(x)\,dx.\)

Hint

\(\cos^2(x)=\frac{1}{2}+\frac{1}{2}\cos(2x)\)

Answer

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

The general process for integrating products of powers of \(\sin(x)\) and \(\cos(x)\) boils down to making a \( u \)-substitution (for either \( \sin(x) \) or \( \cos(x) \)) so that the differential, \( du \), "steals" the odd power off of the other trigonometric function. You then rewrite the entire integrand in terms of \( u \) using the Pythagorean Identity. If there are no odd powers to begin with, then you must use the Power Reduction Formulas. Rather than writing out a "Problem-Solving Strategy" (which is too bulky and algorithmic), we showcase the process through examples.

Example \(\PageIndex{4}\): Integrating \( \cos^j(x)\sin^k(x)\) where \(k\) is Odd

Evaluate \(\displaystyle \int \cos^8(x)\sin^5(x)\,dx.\)

Solution

We would prefer to have all powers even (this gives us the ability to use identities from Trigonometry). Therefore, we let \( u = \cos(x) \) so that \( du = -\sin(x)\,dx \) "steals" the odd power of the sine function away. Thus,

\[ \begin{array}{rclr}
\displaystyle \int \cos^8(x)\sin^5(x)\,dx & = & \displaystyle \int \cos^8(x) \sin^4(x)\sin(x)\,dx & \left( \text{Break off }\sin(x) \right) \\
& = & \displaystyle \int \cos^8(x)(\sin^2(x))^2\sin(x)\,dx & \left(\text{Rewrite }\sin^4(x)=(\sin^2(x))^2\right) \\
& = & \displaystyle \int \cos^8(x)(1−\cos^2(x))^2\sin(x)\,dx & \left(\text{Substitute }\sin^2(x)=1−\cos^2(x)\right) \\
& = & \displaystyle \int u^8(1−u^2)^2(−du) & \left(\text{Let }u=\cos(x)\text{ and }du=−\sin(x)\,dx\right) \\
& = & \displaystyle \int (−u^8+2u^{10}−u^{12})\,du & \\
& = & −\dfrac{1}{9}u^9+\dfrac{2}{11}u^{11}−\dfrac{1}{13}u^{13}+C & \\
& = & −\dfrac{1}{9}\cos^9(x) + \dfrac{2}{11}\cos^{11}(x) − \frac{1}{13}\cos^{13}(x)+C & \\
\end{array}\nonumber \]

Exercise \(\PageIndex{4}\)

Evaluate \(\displaystyle \int \cos^3(x)\,dx.\)

Hint

What substitution would "steal" the odd power away from the cosine function?

Answer

\(\displaystyle \int \cos^3(x)\,dx = \sin(x)−\frac{1}{3}\sin^3(x)+C\)

Example \(\PageIndex{5}\): Integrating \( \cos^j(x)\sin^k(x)\) where \(k\) and \(j\) are Even

Evaluate \(\displaystyle \int \sin^4(x)\,dx.\)

Solution

Since the power on both sine and cosine are even (4 and 0, respectively), any \( u \)-substitution we attempt will result in "stealing" a power off of sine (or cosine), which will result in a trigonometric function not equal to \( u \) having an odd power. This will leave us stuck. Thus, we must resort to using the Power Reduction Formulas. Thus,

\[ \begin{array}{rclr}
\displaystyle \int \sin^4(x)\,dx & = & \int \left(\sin^2(x)\right)^2\,dx & \\
& = & \displaystyle \int \left(\dfrac{1}{2}−\dfrac{1}{2}\cos(2x)\right)^2\,dx & \\
& = & \displaystyle \int \left(\dfrac{1}{4}−\dfrac{1}{2}\cos(2x)+\dfrac{1}{4}\cos^2(2x)\right)\,dx & \\
& = & \displaystyle \int \left(\dfrac{1}{4}−\dfrac{1}{2}\cos(2x)+\dfrac{1}{4}\left(\dfrac{1}{2}+\dfrac{1}{2}\cos(4x)\right)\right)\,dx & \left(\text{Since }\cos^2(2x)\text{ has an even power, substitute }\cos^2(2x)=\frac{1}{2}+\frac{1}{2}\cos(4x)\right) \\
& = & \displaystyle \int \left(\dfrac{3}{8}−\dfrac{1}{2}\cos(2x)+\dfrac{1}{8}\cos(4x)\right)\,dx & \\
& = & \dfrac{3}{8}x−\dfrac{1}{4}\sin(2x)+\dfrac{1}{32}\sin(4x)+C & \\
\end{array} \nonumber \]

Exercise \(\PageIndex{5}\)

Evaluate \(\displaystyle \int \cos^2(3x)\,dx.\)

Hint

Since all powers of sine and cosine are even, we must use Trigonometry (via Power Reduction Formulas) prior to Calculus.

Answer

\(\displaystyle \int \cos^2(3x)\,dx = \frac{1}{2}x+\frac{1}{12}\sin(6x)+C\)

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 \(\sin(ax), \sin(bx), \cos(ax),\) and \(\cos(bx).\) These integrals are evaluated by applying the Product-to-Sum Formulas from Trigonometry. These identities are listed below because no student should be made to memorize them.

Theorem: Product-to-Sum Formulas

\[\sin(ax)\sin(bx)=\frac{1}{2}\cos((a−b)x)−\frac{1}{2}\cos((a+b)x) \nonumber \]

\[\sin(ax)\cos(bx)=\frac{1}{2}\sin((a−b)x)+\frac{1}{2}\sin((a+b)x) \nonumber \]

\[\cos(ax)\cos(bx)=\frac{1}{2}\cos((a−b)x)+\frac{1}{2}\cos((a+b)x) \nonumber \]

Example \(\PageIndex{6}\): Evaluating \( \int \sin(ax)\cos(bx)\,dx\)

Evaluate \(\displaystyle \int \sin(5x)\cos(3x)\,dx.\)

Solution

Apply the identity \(\sin(5x)\cos(3x)=\frac{1}{2}\sin(2x)+\frac{1}{2}\sin(8x).\) Thus,

\[\int \sin(5x)\cos(3x)\,dx= \int \dfrac{1}{2}\sin(2x)+\dfrac{1}{2}\sin(8x)\,dx=−\dfrac{1}{4}\cos(2x)−\dfrac{1}{16}\cos(8x)+C.\nonumber \]

Exercise \(\PageIndex{6}\)

Evaluate \(\displaystyle \int \cos(6x)\cos(5x)\,dx.\)

Hint

Substitute \(\cos(6x)\cos(5x)=\frac{1}{2}\cos x+\frac{1}{2}\cos(11x).\)

Answer

\(\displaystyle \int \cos(6x)\cos(5x)\,dx = \frac{1}{2}\sin x+\frac{1}{22}\sin(11x)+C\)

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Integrating Products and Powers of \(\tan(x)\) and \(\sec(x)\)

Before discussing the integration of products and powers of \(\tan(x)\) and \(\sec(x)\), we need to introduce the antiderivative of the secant.

Theorem: \( \displaystyle \int \sec(x) \, dx \) and \( \displaystyle \int \csc(x) \, dx \)

\[ \int \sec(x) \, dx = \ln|\sec(x) + \tan(x)| + C \nonumber \]

and

\[ \int \csc(x) \, dx = -\ln|\csc(x) + \cot(x)| + C \nonumber \]

Proof

\[ \begin{array}{rclr}
\displaystyle \int \sec(x) \, dx & = & \displaystyle \int \dfrac{\sec(x) \left( \sec(x) + \tan(x) \right)}{\sec(x) + \tan(x)} \, dx & \left( \text{Multiply numerator and denominator by }\sec(x) + \tan(x) \right)\\
& = & \displaystyle \int \dfrac{\sec^2(x) + \sec(x) \tan(x)}{\sec(x) + \tan(x)} \, dx & \\
& = & \displaystyle \int \dfrac{1}{u} \, du & \left( \text{Let }u = \sec(x) + \tan(x) \implies du = (\sec(x) \tan(x) + \sec^2(x))\, dx \right) \\
& = & \ln|u| + C & \\
& = & \ln|\sec(x) + \tan(x)| + C & \left( \text{Resubstitute} \right) \\
\end{array} \nonumber \]The proof for the integral of the cosecant is generally left as a homework exercise.

Now we can collect the integrals involving \(\tan(x)\) and \(\sec(x)\) we have learned:

• \(\displaystyle \int \sec^2(x)\,dx=\tan(x)+C\)
• \(\displaystyle \int \sec(x)\tan(x)\,dx=\sec(x)+C\)
• \(\displaystyle \int \tan(x)\,dx=\ln|\sec(x)|+C\)
• \(\displaystyle \int \sec(x)\,dx=\ln|\sec(x)+\tan(x)|+C\)

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 \(\displaystyle \int \tan^j(x)\sec^2(x)\,dx\) or \(\displaystyle \int \sec^j(x)\tan(x)\,dx\). In the former case, we can see that allowing \( u = \tan(x) \) would cause \( du \) to "steal" the \( \sec^2(x) \, dx \). In the latter case, allowing \( u = \sec(x) \) causes \( du \) to "steal" a single power of \( \sec(x) \) along with the \( \tan(x) \, dx \). In both cases, the integral simplifies nicely.

Let's see this "in action."

Example \(\PageIndex{7}\): Evaluating \( \int \sec^j(x)\tan(x)\,dx\)

Evaluate \(\displaystyle \int \sec^5(x)\tan(x)\,dx.\)

Solution

Choosing to let \( u = \tan(x) \) would result in \( du = \sec^2(x) \, dx \) "stealing" two powers off the secant function. This would result in an odd power remaining on secant (a power of 3, in fact). This is not desirable. Therefore, we let \( u = \sec(x) \) so that \( du = \sec(x) \tan(x) \, dx \). This substitution would "steal" a single power from the tangent, leaving an even power (0, in fact) on tangent, which is what we want.

Thus,

\[ \begin{array}{rclr}
\displaystyle \int \sec^5(x)\tan(x)\,dx & = & \displaystyle \int \sec^4(x)\sec(x)\tan(x) \,dx & \\
& = & \displaystyle \int u^4\,du & \left(\text{Let } u=\sec(x)\text{, then }du=\sec(x)\tan(x) \,dx\right) \\
& = & \displaystyle \dfrac{1}{5}u^5+C & \\
& = & \dfrac{1}{5}\sec^5(x)+C & \\
\end{array} \nonumber \]

Exercise \(\PageIndex{7}\)

Evaluate \(\displaystyle \int \tan^5(x)\sec^2(x)\,dx.\)

Hint

Let \(u=\tan(x)\) and \(du=\sec^2(x) \, dx\).

Answer

\(\displaystyle \int \tan^5(x)\sec^2(x)\,dx = \frac{1}{6}\tan^6(x)+C\)

When working with products and powers of the sine and cosine, the "simple" rule of thumb was to make a \( u \)-substitution so that the "non-\( u \)" function ended with an even power. We then use the Pythagorean Identity to rewrite the integrand completely in terms of \( u \). The only except to this strategy was if both sine and cosine have even powers. In that case, we start by using the Power Reduction Formulas.

When dealing with products and powers of the secant and the tangent, things are not so clean. As such, writing out a "Problem-Solving Strategy" is helpful, but still not as good as experience.

Problem-Solving Strategy: Integrating \(\displaystyle \int \tan^k(x)\sec^j(x)\,dx\)

To integrate \(\displaystyle \int \tan^k(x)\sec^j(x)\,dx,\) use the following strategies:

1. 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 \(du=\sec^2(x)\,dx\).
2. 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 \(du=\sec(x)\tan(x)\,dx.\) (Note: If \(j\) is even and \(k\) is odd, then either strategy 1 or strategy 2 may be used.)
3. 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.
4. 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 \(\sec(x).\)

Caution

The preceding "Problem-Solving Strategy" is complex, but there really is no way around it. As was stated in Section 2.1, integration techniques are mostly heuristic methods and its best to learn them by using them on a lot of problems - experience is a much better teacher than memorizing the preceding strategy.

One strategy that I almost always try is to make a \( u \)-substitution that will result in the non-\( u \) function having an even power. This strategy still works about 75% of the time with products of powers of secant and tangent functions; however, it does not always work. Still, it's a good starting point.

Example \(\PageIndex{8}\): Integrating \( \tan^k(x)\sec^j(x)\) when \(j\) is Even

Evaluate \(\displaystyle \int \tan^6(x)\sec^4(x)\,dx.\)

Solution

Both powers on tangent and secant are already even, so whatever \( u \)-substitution (if any) that we perform, we would like to keep those even powers there (this allows us to use our trigonometric identities, if needed). To keep the powers even after a \( u \)-substitution, let's set \( u = \tan(x) \) so that \( du = \sec^2(x) \, dx \) "steals" two powers off the secant function (keeping the power even in the end). Thus,

\[\begin{array}{rclr}
\displaystyle \int \tan^6(x)\sec^4(x)\,dx & = & \displaystyle \int \tan^6(x)(\tan^2(x)+1)\sec^2(x)\,dx & \\
& = & \displaystyle \int u^6(u^2+1)\,du & \left(\text{Let }u=\tan(x)\text{ and }du=\sec^2(x) dx\right) \\
& = & \displaystyle \int (u^8+u^6)\,du & \\
& = & \dfrac{1}{9}u^9+\dfrac{1}{7}u^7+C & \\
& = & \dfrac{1}{9}\tan^9(x)+\dfrac{1}{7}\tan^7(x)+C. & \\
\end{array} \nonumber \]

Example \(\PageIndex{9}\): Integrating \( \tan^k(x)\sec^j(x)\) when \(k\) is Odd

Evaluate \(\displaystyle \int \tan^5(x)\sec^3(x)\,dx.\)

Solution

Again, we start with the goal to force even powers on the non-\( u \) function. If we let \( u = \sec(x) \), then \( du = \sec(x) \tan(x) \, dx \) will "steal" enough power so that the tangent (the non-\( u \) function) is left with an even power. Thus,

\[\begin{array}{rclr}
\displaystyle \int \tan^5(x)\sec^3(x)\,dx & = & \displaystyle \int \tan^4(x)\sec^2(x)\sec(x)\tan(x) \, dx & \\
& = & \displaystyle \int (\tan^2(x))^2\sec^2(x)\sec(x)\tan(x)\,dx & \\
& = & \displaystyle \int (\sec^2(x)−1)^2\sec^2(x)\sec(x)\tan(x)\,dx & \\
& = & \displaystyle \int (u^2−1)^2u^2du & \left(\text{Let }u=\sec(x)\text{ and }du=\sec(x)\tan(x)\,dx\right) \\
& = & \displaystyle \int (u^6−2u^4+u^2)du & \\
& = & \dfrac{1}{7}u^7−\dfrac{2}{5}u^5+\dfrac{1}{3}u^3+C & \\
& = & \dfrac{1}{7}\sec^7(x)−\dfrac{2}{5}\sec^5(x)+\dfrac{1}{3}\sec^3(x)+C & \\
\end{array} \nonumber \]

Example \(\PageIndex{10}\): Integrating \( \tan^k(x)\) where \(k\) is Odd and \(k \geq 3\)

Evaluate \(\displaystyle \int \tan^3(x)\,dx.\)

Solution

We could probably stare at this for a while and think, "What kind of substitution would lead to an even power on that tangent?" The answer is, in the form this integral is written, no substitution will help! This is where we need to get a bit creative. If we recognize that

\[ \tan^3(x) = \tan(x) \tan^2(x) = \tan(x) \left( \sec^2(x) - 1) \right) = \tan(x) \sec^2(x) - \tan(x), \nonumber \]

we can make some progress.

\[ \begin{array}{rcl}
\displaystyle \int \tan^3(x)\,dx & = & \displaystyle \int (\tan(x)\sec^2(x)−\tan(x))\,dx \\
& = & \displaystyle \int \tan(x)\sec^2(x)\,dx − \displaystyle \int \tan(x)\,dx \\
& = & \dfrac{1}{2}\tan^2(x) − \ln |\sec(x)|+C. \\
\end{array} \nonumber \]

For the first integral, we used the substitution \(u=\tan(x).\) For the second integral, we used the formula.

Example \(\PageIndex{11}\): Integrating \( \sec^3(x)\)

Evaluate \(\displaystyle \int \sec^3(x)\,dx\).

Solution

This integral is dangerously popular in some texts!

Just like Example \( \PageIndex{10} \), any choice of substitution will lead nowhere. For example, letting \( u = \tan(x) \) will definitely "steal" two powers off the secant (via \( du = \sec^2(x) \, dx \)); however, secant would be left with an odd power - we don't want that. Letting \( u = \sec(x) \) would steal a power off tangent (via \( du = \sec(x) \tan(x) \, dx \)), leaving \( \left( \tan(x) \right)^{-1} \); however, we would need that power on tangent to be even, and that's not going to happen.

This is a great example of a problem where we need to reach into our "Bag of Integration Techniques" to try something different. So far, our integration techniques include the Substitution Method and Integration by Parts. Since there's not much to be "substituted" here, let's try Integration by Parts.

We would like to rewrite the integrand as the product of two functions (which is not technically always needed for Integration by Parts, but it is helpful here):

\[ \int \sec^3(x) \, dx = \int \sec(x) \sec^2(x) \, dx. \nonumber \]

We definitely can integrate \( sec^2(x) \), so we choose \( dv = \sec^2(x) \, dx \) and \( u = \sec(x) \).

\( u = \sec(x) \)	\( dv = \sec^2(x) \, dx \)
\( du = \sec(x) \tan(x) \, dx \)	\( v = \tan(x) \)

Thus,

\[\begin{array}{rcl}
\displaystyle \int \sec^3(x)\,dx & = & \sec(x)\tan(x) − \displaystyle \int \tan(x)\sec(x)\tan(x)\,dx \\
& = & \sec(x)\tan(x) − \displaystyle \int \tan^2(x)\sec(x)\,dx \\
& = & \sec(x)\tan(x) − \displaystyle \int (\sec^2(x)−1)\sec(x)\,dx \\
& = & \sec(x)\tan(x) + \displaystyle \int \sec(x)\,dx − \displaystyle \int \sec^3(x)\,dx \\
& = & \sec(x)\tan(x) + \ln|\sec(x)+\tan(x)| − \displaystyle \int \sec^3(x)\,dx. \\
\end{array} \nonumber \]

We now have

\[ \int \sec^3(x)\,dx=\sec(x)\tan(x)+\ln|\sec(x)+\tan(x)|− \int \sec^3(x)\,dx.\nonumber \]

Since the integral \(\displaystyle \int \sec^3(x)\,dx\) has reappeared on the right-hand side, we can solve for \(\displaystyle \int \sec^3(x)\,dx\) by adding it to both sides. In doing so, we obtain

\[2 \int \sec^3(x)\,dx = \sec(x)\tan(x) + \ln|\sec(x)+\tan(x)|.\nonumber \]

Dividing by 2, we arrive at

\[ \int \sec^3(x)\,dx = \dfrac{1}{2}\sec(x)\tan(x)+\dfrac{1}{2}\ln|\sec(x)+\tan(x)|+C.\nonumber \]

Note

While I love the derivation of this integral (and you could be expected to derive it on an exam), I cannot help but mention the cool pattern it generates. The integral of the cube of the secant is one-half the sum of the derivative of the secant and the integral of the secant.

Exercise \(\PageIndex{11}\)

Evaluate \(\displaystyle \int \tan^3(x)\sec^7(x)\,dx\).

Hint

Use Example \(\PageIndex{9}\) as a guide.

Answer

\(\displaystyle \int \tan^3(x)\sec^7(x)\,dx = \frac{1}{9}\sec^9(x)−\frac{1}{7}\sec^7(x)+C\)

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Reduction Formulas

Evaluating \(\displaystyle \int \sec^n(x)\,dx\) for values of \(n\) where \(n\) is odd requires Integration by Parts. In addition, we must also know the value of \(\displaystyle \int \sec^{n−2}(x)\,dx\) to evaluate \(\displaystyle \int \sec^n(x) \,dx\). In addition, the evaluation of \(\displaystyle \int \tan^n(x)\,dx\) also requires being able to integrate \(\displaystyle \int \tan^{n−2}(x)\,dx\). To make the process easier, we can derive and apply the following Power Reduction Formulas for Integration. 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 \(\tan(x)\).

Theorem: Power Reduction Formulas for \( \int \sec^n(x)\,dx\) and \( \int \tan^n(x)\,dx\)

\[ \int \sec^n(x)\,dx=\dfrac{1}{n−1}\sec^{n−2}(x)\tan(x)+\dfrac{n−2}{n−1} \int \sec^{n−2}(x)\,dx \nonumber \]

and

\[ \int \tan^n(x)\,dx=\dfrac{1}{n−1}\tan^{n−1}(x) − \int \tan^{n−2}(x)\,dx. \nonumber \]

Note

The Power Reduction Formula for Integrating Powers of Secant may be verified by applying Integration by Parts. The Power Reduction Formula for Integrating Powers of Tangent may be verified by following the strategy outlined for integrating odd powers of \(\tan(x).\)

Example \(\PageIndex{12}\): Revisiting \( \int \sec^3(x)\,dx\)

Apply a reduction formula to evaluate \(\displaystyle \int \sec^3(x)\,dx.\)

Solution

By applying the first reduction formula, we obtain

\[\begin{array}{rcl}
\displaystyle \int \sec^3(x)\,dx & = & \dfrac{1}{2}\sec(x)\tan(x)+\dfrac{1}{2} \displaystyle \int \sec(x)\,dx \\
& = & \dfrac{1}{2}\sec(x)\tan(x)+\frac{1}{2}\ln|\sec(x)+\tan(x)|+C. \\
\end{array} \nonumber \]

Example \(\PageIndex{13}\): Using a Power Reduction Formula for Integration

Evaluate \(\displaystyle \int \tan^4(x)\,dx.\)

Solution

Applying the Power Reduction Formula for \( \int \tan^4(x)\,dx\) we have

\[\begin{array}{rcl}
\displaystyle \int \tan^4(x)\,dx & = & \dfrac{1}{3}\tan^3(x)− \displaystyle \int \tan^2(x)\,dx\\
& = & \dfrac{1}{3}\tan^3(x)−(\tan(x)− \displaystyle \int \tan^0(x)\,dx) \\
& = & \dfrac{1}{3}\tan^3(x)−\tan(x)+ \displaystyle \int 1\,dx \\
& = & \dfrac{1}{3}\tan^3(x)−\tan(x)+x+C \\
\end{array} \nonumber \]

Exercise \(\PageIndex{9}\)

Apply the Power Reduction Formula to \(\displaystyle \int \sec^5(x)\,dx.\)

Hint

Use the Power Reduction Formula for Integrals of Powers of Secant and let \(n=5.\)

Answer

\(\displaystyle \int \sec^5(x)\,dx=\frac{1}{4}\sec^3(x)\tan(x)+\frac{3}{4} \int \sec^3(x) \, dx\)

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Key Concepts

Integrals of trigonometric functions can be evaluated by the use of various strategies. These strategies include

1. Applying trigonometric identities to rewrite the integral so that it may be evaluated by \(u\)-substitution
2. Using Integration by Parts
3. Applying trigonometric identities to rewrite products of sines and cosines with different arguments as the sum of individual sine and cosine functions
4. Applying reduction formulas

Key Equations

To integrate products involving \(\sin(ax), \,\sin(bx), \,\cos(ax),\) and \(\cos(bx),\) use the substitutions.

• Sine Products

\(\sin(ax)\sin(bx)=\frac{1}{2}\cos((a−b)x)−\frac{1}{2}\cos((a+b)x)\)

• Sine and Cosine Products

\(\sin(ax)\cos(bx)=\frac{1}{2}\sin((a−b)x)+\frac{1}{2}\sin((a+b)x)\)

• Cosine Products

\(\cos(ax)\cos(bx)=\frac{1}{2}\cos((a−b)x)+\frac{1}{2}\cos((a+b)x)\)

• Power Reduction Formula

\(\displaystyle \int \sec^nx\,dx=\frac{1}{n−1}\sec^{n−2}x \tan x+\frac{n−2}{n−1} \int \sec^{n−2}x\,dx\)

• Power Reduction Formula

\(\displaystyle \int \tan^nx\,dx=\frac{1}{n−1}\tan^{n−1}x− \int \tan^{n−2}x\,dx\)

Glossary

power reduction formula

a rule that allows an integral of a power of a trigonometric function to be exchanged for an integral involving a lower power

trigonometric integral

an integral involving powers and products of trigonometric functions