Trigonometric integrals

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

We can use substitution and trigonometric identities to find antiderivatives of certain types of trigonometric functions.

In this section, we will study how to integrate certain types of trigonometric functions. We begin by giving the antiderivatives of the six basic trigonometric functions:

Antiderivatives of other trigonometric functions

Antiderivatives of trigonometric functions

$\int \sin (x) dx = - \cos (x)+C$
$\int \cos (x) dx = \sin (x)+C$
$\int \tan (x) dx = \ln |\sec (x)| + C$
$\int \cot (x) dx = -\ln |\csc (x)| + C$
$\int \sec (x) dx = \ln |\sec (x) + \tan (x)| + C$
$\int \csc (x) dx = -\ln |\csc (x) + \cot (x)| + C$

We are able to establish the antiderivatives of $\sin (x)$ and $\cos (x)$ easily from the formulas for their derivatives, but how do we establish the other formulas?

Compute $\int \tan (x) dx$ .

This can be computed with a substitution once we recognize:

$\int \tan (x) dx = \int \frac {\sin (x)}{\cos (x)} dx$

By setting $u = \answer [given]{\cos (x)}$ , we find $du = \answer [given]{-\sin (x)} dx$ and thus:

$\int \tan (x) dx = \int \frac {\sin (x)}{\cos (x)} dx =\int \answer {- \frac {1}{u}} du =-\ln |u|+C = -\ln |\cos (x)|+C$

Using the properties of logarithms, followed by the fact that $\sec (x) = \frac {1}{\cos (x)}$ gives:

$-\ln |\cos (x)| = \ln \left | (\cos (x))^{-1} \right | = \ln \left | \frac {1}{\cos (x)} \right | = \ln |\sec (x)|$

Compute $\int \sec (x) dx$ .

It may not be clear why, but let’s multiply the top and bottom by $\sec (x) + \tan (x)$ .

$\int \sec (x) dx= \int \sec (x) \left ( \frac { \sec (x) + \tan (x)}{\sec (x)+\tan (x)} \right ) dx = \int \frac { \sec ^{2}(x) + \sec (x)\tan (x)}{\sec (x)+\tan (x)} dx$

Now we see the rationale behind multiplying top and bottom by $\sec (x) + \tan (x)$ . The derivative of the denomiator of the integrand is precisely the numerator.

If we let $u=\sec (x)+\tan (x)$ then $du=\sec (x)\tan (x)+\sec ^{2}(x) dx$ and thus our integral becomes

$\int \sec (x) dx = \int \frac {1}{u} du=\ln |u|+C=\ln |\sec (x)+\tan (x)|+C$

The formulas for the antiderivatives of $\csc (x)$ and $\cot (x)$ are found analogously.

Integrals involving products of $\sin (x)$ and $\cos (x)$

We begin with a simple example:

Compute $\int \sin ^{2}(x) \cos (x) dx$ .

Here we recognize that $\cos (x)$ is the derivative of $\sin (x)$ so substitution is a good method to use. Let $u=\sin (x)$ . Then $du=\cos (x) dx$ and:

$\int \sin ^{2}(x) \cos (x) dx = \int u^{2} dx=\frac {u^{3}}{3}+C=\frac { \sin ^{3}(x)}{3}+C$

In this example, we saw that a substitution was quite helpful. Earlier, we saw that there were two common instances in which substitutions could be helpful to compute antiderivatives:

Once we change variables, the remaining expressions in the old variable are part of the differential in the new one. For instance, consider:
$\int 2x \cos \left (x^2\right ) dx$ By setting $u=x^2$ , $du = 2x dx$ , so the $2x dx$ in the original integral is part of the new differential $du$ and:
$\int 2x \cos \left (x^2\right ) dx = \int \cos (u) du$
Once we change variables, it is not complicated to write the remaining old variables in terms of the new one. For instance, consider:
$\int 2x^3 \cos \left (x^2\right ) dx$ By setting $u=x^2$ , $du = 2x dx$ , so the $2x dx$ in the original integral is part of the new differential $du$ and:
$\int 2x^3 \cos \left (x^2\right ) dx = \int x^2 \cos (u) du = \int u \cos (u) du$
where we have used the fact that $u=x^2$ in the last step above. Note that we can do this last integral using integration by parts!

This highlights a strategy that we will employ in this section; try to let one of the trigonometric functions be our new variable, and express the remaining expressions easily in terms of it.

Compute $\int \sin ^{4}(x) \cos ^{3}(x) dx$ .

Per the previous discussion, we have two options: we can let $u= \sin (x)$ or $u=\cos (x)$ . If we let $u = \sin (x)$ , we will have to reserve a copy of $\cos (x)$ for the differential. This leaves two powers of $\cos (x)$ left. One of our fundamental trig identities, the Pythagorean identity: $\cos ^{2}(x) + \sin ^{2}(x)=1$ can be used to covert between even powers of sines and cosines, so this will be a fruitful approach:

We can use this identity to rewrite $\cos ^{2}(x)=1-\sin ^{2}(x)$ in our integral so that it becomes: $\int \sin ^{4}(x) (1-\sin ^{2}(x)) \cos (x) dx$ Now the same strategy we used in the first example of this section will work. We can do the substitution $u=\sin (x)$ . We find $du= \answer [given]{\cos (x)} dx$ , so that the integral becomes:

$\int \sin ^{4}(x) (1-\sin ^{2}(x)) \cos (x) dx = \int u^{4}(1-u^{2}) du$

Finishing this integral we get

$\int u^{4}(1-u^{2}) du= \int u^{4}-u^{6} du = \answer [given]{\frac {1}{5}u^5 - \frac {1}{7}u^7} + C =\frac {1}{5}\sin ^{5}(x) - \frac {1}{7} \sin ^{7}(x) + C$

Note that had we let $u = \cos (x)$ , we will have to reserve a copy of $\sin (x)$ for the differential. This leaves three powers of $\sin (x)$ left! We certainly can use the Pythagorean identity from before to write:

$\sin ^3(x) = \left (\sin ^2(x)\right )^{3/2} = \left (1-\cos ^2(x)\right )^{3/2} = (1-u^2)^{3/2}$

If $u = \cos (x)$ , we find $du = -\sin (x) dx$ , and from the above we can write the original integral as:

$\int \sin ^{4}(x) \cos ^{3}(x) dx = \int \sin ^4(x) u^3 \frac {du}{-\sin (x)} = -\int \sin ^3(x) \cdot u^3 du = -\int (1-u^2)^{3/2} u^3 du$ This integral can be evaluated with another substitution $v = 1-u^2$ and some algebra, but it is certainly more time-consuming than the first approach!

As a reminder, the integrand should only display the variable of integration once the substitution has been completed; $x$ and $u$ cannot appear in the same integral in the final step since they depend on each other! The previous example was typeset the way that it was in order to exhibit the simplification in a clear way.

We can make a useful observation from this example. Our previous integral involved powers of $\sin (x)$ and $\cos (x)$ . We peeled off a copy of the derivative of what we wanted to use for $u$ . If this leaves an even number of the other trigonometric function left, we can use the Pythagorean identity $\cos ^{2}(x)+ \sin ^{2}(x) =1$ to write the remaining powers of the other trigonometric function in terms of $u$ .

In the preceding example, the power of cosine was odd, and for other examples, as long as the power of $\cos (x)$ in the original integral is odd, then we can peel off a $\cos (x)$ term to be used for $du$ and then let $u=\sin (x)$ to express the integral in terms of $u$ . This general approach is stated in more detail in the exercises, but for now let’s try another example.

Compute $\int \sin ^{5}(x) \cos ^2(x) dx$ .

If we try $u=\sin (x)$ , we find that we need a factor of $\cos (x)$ for the differential, which leaves 1 power of $\cos (x)$ left, so using the Pythagorean identity would produce an unpleasant integral. However, if we try $u=\cos (x)$ , we find that we need a factor of $\sin (x)$ for the differential, which leaves four powers of $\sin (x)$ left. We can rewrite this suggestively as:

$\begin{align*} \int &\sin^4(x) \cos^2(x) \sin(x) dx \\ &= \int (\sin^2(x))^2 \cos^2(x) \sin(x) dx \end{align*}$ Now using the Pythagorean identity we can rewrite everything in terms of $\cos (x)$ , but reserve the one power of $\sin (x)$ to help with the differential in an eventual variable substitution $u=\cos (x)$ :

Making the substitution

$\begin{align*} u &= \cos(x)\\ du &=\answer[given]{-\sin(x)} dx \end{align*}$ yields $\int \answer [given]{-(1-u^2)^2 u^2} du$ This is a polynomial in $u$ , and so we can easily finish the integration by just expanding the polynomial out and integrating term-by-term. Write with me $\begin{align*} \int {-(1-u^2)^2 u^2} du &= \int -(1-2u^2+u^4)u^2du\\ &= \int-u^6 + 2u^4-u^2 du\\ &=\answer[given]{-\frac{1}{7}u^7 + \frac{2}{5}u^5 -\frac{1}{3} u^3}+C \end{align*}$ Substituting $u = \cos (x)$ we find our final answer: $-\frac {1}{7}\cos ^7(x) + \frac {2}{5}\cos ^5(x) - \frac {1}{3} \cos ^3(x)+C$

Sometimes we need to massage our function into the correct form:

Compute $\int \sin ^5(x) dx$ .

We start by separating a power of sine for an eventual substitution $\begin{align*} \int \sin^5 xdx &=\int \sin^4(x) \sin(x) dx\\ &=\int (\sin^2(x))^2 \sin(x) dx \end{align*}$ Here is what we’re thinking:

Now set $u=\cos x$ , $du=-\sin xdx$ :

$\begin{align*} \int \sin x (1-\cos^2 x)^2dx&=\int \answer[given]{-(1-u^2)^2} du \\ &=\int -(1-2u^2+u^4)du \\ &=\answer[given]{-u+\frac{2}{3}u^3-\frac{1}{5}u^5}+C \end{align*}$ Hence our final answer is: $\answer [given]{-\cos x+\frac {2}{3}\cos ^3 x-\frac {1}{5}\cos ^5x}+C$

Another strategy

Thinking again about powers of cosine and sine, what do we do when both powers are even? Our strategy above is sunk, since peeling-off one power leaves an odd power, and we cannot use the Pythagorean identity to rewrite this in a nice way. Instead, we will use the “power-reduction formulas”:

Cosine Power-Reduction: : $\cos ^2(\theta )= \frac {1}{2}+\frac {1}{2}\cos (2\theta )$
Sine Power-Reduction: : $\sin ^2(\theta ) = \frac {1}{2}-\frac {1}{2}\cos (2\theta )$

Using the above formula we find $\sin ^2(2x) = \frac {1}{2}-\frac {1}{2} \cos (\answer {4x})$ .

The designation “power reduction formula” sounds quite impressive, but they result from a simple rearrangement of the double angle formula for $\cos (2\theta )$ . Indeed if we want to obtain the power reduction formula for $\cos ^2(\theta )$ : $\begin{align*} \cos(2\theta) &= \cos^2(\theta)-\sin^2(\theta) \\ \cos(2\theta) &= \cos^2(\theta) - (1-\cos^2(\theta)) \\ \cos(2\theta) &= 2\cos^2(\theta)-1 \end{align*}$ Solving the above equation for $\cos ^2(\theta )$ gives: $\cos ^2(\theta ) = \frac {1}{2} +\frac {1}{2} \cos (2\theta )$ .

In this case, it is (usually) critical to apply the power-reduction formulas to every instance of cosine and sine appearing in the integrand.

First consider a simple example:

Compute $\int \cos ^{2}(x) dx$ .

The strategy of setting $u=\sin (x)$ or $u=\cos (x)$ will not produce an integral that involves an even number of the other type of trigonometric function. Instead, we use the power reduction formula $\cos ^{2}(x)=\frac {1+\cos (2x)}{2}$ to transform the integral into a form we can directly integrate.

$\int \cos ^{2} (x) dx= \frac {1}{2} \int 1+\cos (2x) dx = \frac {1}{2} \left ( x+ \frac { \sin (2x)}{2} \right ) + C$

Compute $\int \sin ^2x\cos ^2xdx$ .

The strategy of setting $u=\sin (x)$ or $u=\cos (x)$ will not produce an integral that involves an even number of the other type of trigonometric function. Use both power-reduction formulas to find: $\int \sin ^2x\cos ^2xdx=\int \left ( \frac {1-\cos (2x)}{2}\right ) \left (\frac {1+\cos (2x)}{2}\right ) dx$ From here, we can try to simplify further by multiplying out the integrand. We may end up having to use the power reduction formulas again to reduce the powers that appear when we multiply out the terms. In this case $\begin{align*} \int \left(\frac{1-\cos(2x)}{2}\right) \left( \frac{1+\cos(2x)}{2}\right) dx &= \frac{1}{4} \int 1-\cos^2(2x) dx\\ &= \frac{1}{4} \int \sin^2(2x) dx \end{align*}$

Since this expression has only even powers of sine, we must use the same strategy again, and employ a power-reduction formula: $\sin ^2(2x) = \frac {1-\cos (\answer [given]{4x})}{2}$ so we find $\frac {1}{8}\int 1-\cos (4x) dx = \answer [given]{\frac {1}{8}(x - \frac {1}{4}\sin (4x))} + C$

Working with products of powers of $\sec (x)$ and $\tan (x)$

The same logic used in the preceding examples can be used to find antiderivatives involving products of $\sec (x)$ and $\tan (x)$ . Indeed, the Pythagorean identity has two other forms: $\sec ^2(x) = 1 + \tan ^2(x) \qquad \text {and}\qquad \tan ^2(x) = \sec ^2(x) - 1$ where the first identity we find by dividing the Pythagorean identity by $\cos ^2(x)$ and the second we find by subtracting 1 from both sides of the first identity.

The derivatives of these functions also can be expressed in terms of each other:

$\frac {d}{dx} \left [\tan (x)\right ] = \sec ^2(x) \qquad \text {and}\qquad \frac {d}{dx} \left [\sec (x)\right ] = \sec (x)\tan (x)$

It is worth seeing several examples where we use these identities. We proceed in a fashion similar to the procedures we saw earlier.

Compute $\int \tan ^3(x) \sec ^4(x) dx$ .

Let’s think about this a bit first. We recall that the derivative of $\tan (x)$ is $\sec ^{2}(x)$ . This suggests that we might want to try $u=\tan (x)$ which requires $du=\sec ^{2}(x)$ .

Since the power of secant is even, we can separate off a factor of $\sec ^{2}(x)$ for our $du$ . That leaves an extra factor of $\sec ^{2}(x)$ that we must deal with. Let’s now write out the details.

$\begin{align*} \int &\tan^3(x) \sec^4(x) dx = \int\tan^3(x) \sec^2(x) \sec^2(x) dx \end{align*}$ Now using the identity $1+\tan ^{2}(x)=\sec ^{2}(x)$ we can rewrite everything in terms of $\tan (x)$ , but reserve the one power of $\sec ^2(x)$ to help with the differential in an eventual variable substitution $u=\tan (x)$ :

Making the substitution

$\begin{align*} u &= \tan(x)\\ du &=\answer[given]{\sec^2(x)} dx \end{align*}$ yields $\int \answer [given]{u^3(1+u^2)} du$ This is a polynomial in $u$ , and so we can easily finish the integration by just expanding the polynomial out and integrating term-by-term. We leave it to the intrepid young mathematician to finish this problem.

Compute $\int \tan ^3(x) \sec ^5(x) dx$

Now the power of $\sec (x)$ is not even. If we try to separate off a $\sec ^{2}(x)$ term, the remaining power will be odd and we won’t be able to rewrite it nicely in terms of $\tan (x)$ . We try another strategy. Recall that $\frac {d}{dx} \sec (x)=\sec (x) \tan (x)$

Since the power of tangent is odd and the power of secant is odd, we can pull out a $\sec (x)\tan (x)$ factor and we have

$\begin{align*} \int &\tan^3(x) \sec^5(x) dx \\ &= \int\tan^2(x) \sec^4(x) \sec(x)\tan(x) dx \end{align*}$ Now using the identity $\tan ^{2}(x)=\sec ^{2}(x)-1$ we can rewrite the remaining $\tan ^{2}(x)$ in terms of $\sec (x)$ , but reserve the one power of $\sec (x)\tan (x)$ to help with the differential in an eventual variable substitution $u=\sec (x)$ :

Making the substitution

$\begin{align*} u &= \sec(x)\\ du &=\answer[given]{\sec(x)\tan(x)} dx \end{align*}$ yields $\int \answer [given]{(u^2-1)u^4} du$ This is a polynomial in $u$ , and so we can easily finish the integration by just expanding the polynomial out and integrating term-by-term. We leave it to the interested mathematician to finish this problem.

There is no analogue for the power reduction formulas for sines and cosines, so sometimes, these integrals can become a bit trickier.

Compute $\int \sec ^{4}(x) dx$ .

There are no powers of $\tan (x)$ immediately available here, so letting $u=\tan (x)$ may seem like a bad idea at first. However, by saving a copy of $\sec ^2(x)$ for the differential, we have $2$ powers of $\sec (x)$ left, so let’s give it a try: $\begin{align*} \int \sec^{4}(x) dx &= \int \sec^2(x) \sec^2(x) dx \end{align*}$ Now using the identity $\sec ^{2}(x)=\tan ^{2}(x)+1$ we can write:

Integrating this, we find:

$\begin{align*} \int \sec^{4}(x) dx &= \frac{1}{3}u^3+u +C\\ &= \frac{1}{3} \tan^3(x)+\tan(x)+C \end{align*}$

Comput $\int \sec ^{3}(x) dx$ .

Our previous techniques for handling powers of $\sec (x)$ and $\tan (x)$ relied on having an even power of $\sec (x)$ or an odd power of $\tan (x)$ . Suppose we split off a $\sec ^{2}(x)$ to write the integral as $\int \sec (x) \sec ^{2}(x) dx$ The $\sec ^{2}(x)$ is the antiderivative of $\tan (x)$ but there is no $\tan (x)$ term so $u$ substitution will not work. Since we have a product of two functions where one of the functions can be easily integrated let’s try integration by parts.

Choose $u= \sec (x) \text { and } dv=\sec ^{2}(x) \\$ $du= \sec (x)\tan (x) dx \text { and } v=\tan (x)$

We recall the integration by parts formula $\int u dv=uv-\int v du$ , so we obtain

$\int \sec ^{3}(x) dx=\sec (x)\tan (x) - \int \tan ^{2}(x)\sec (x)dx$

The rightmost integral does not have an odd power of $\tan (x)$ or an even power of $\sec (x)$ but we could convert everything to be in terms of $\sec (x)$ by using the identity $\tan ^{2}(x)=\sec ^{2}(x)-1$ .

The rightmost integral then becomes $\int \tan ^{2}(x)\sec (x) dx=\int (\sec ^{2}(x)-1) \sec (x) dx= \int \sec ^{3}(x)dx - \int \sec (x) dx$

Combining this with our original integral and using the previous example which gave us $\int \sec (x) dx$ we get $\int \sec ^3(x) dx=\sec (x) \tan (x)- \int \sec ^3(x) dx + \ln | \sec (x)+\tan (x) | + C$

Notice that the remaining integral on the right hand side is simply the original integral we were trying to determine. You will remember that we first experienced this phenomenon in the section on integration by parts when we tried to compute $\int e^{x}\cos (x) dx$ . Since the desired unknown quantity appears on the left and the right we simply solve for it and so

$\int \sec ^{3}(x) dx=\frac {1}{2}\left (\sec (x) \tan (x) + \ln | \sec (x) + \tan (x)| \right ) + C$

This last example can be generalized to a reduction formula for $\int \sec ^{n}(x) dx$ . We omit the details but using integration by parts in a similar fashion as above with $u=\sec ^{n-2}(x)$ and $dv=\sec ^{2}(x)$ gives us the following reduction formula:

$\int \sec ^{n}(x) dx= \frac {\tan (x) \sec ^{n-2}(x)}{n-1} + \frac {n-2}{n-1} \int \sec ^{n-2}(x) dx$

We won’t derive them here but note that there are similar reduction formulas that we could derive for integrating powers of other trigonometric functions using integration by parts.

Note, it is also possible to run across trigonometric integrals where none of these techniques immediately apply. If all else fails, one strategy is to convert all trigonometric functions to sines and cosines.

The same strategies can be employed to work with integrals involving powers of cosecants and cotangents since both the Pythagorean identities and the derivatives of these functions behave analogously to those for tangents and secants.

Final thoughts

When we encounter integrals that involve products of complimentary trigonometric functions (sines and cosines, tangents and secants, or cosecants and cotangents), we can employ a general strategy to find the antiderivatives:

Let $u$ be one of the trigonometric functions. Peel off a copy of the derivative for the differential. If this leaves an even number of the other trigonometric function, use the appropriate Pythagorean identity to write the remaining powers of the other trigonometric function in terms of $u$ .
If the integral involves a product of sines and cosines and the above does not work, try using the power reduction formulas.
If the integral involves secants and tangents (or cosecants and cotangents), try either integration by parts or convert everything to sines and cosines.

As usual, practice is necessary to develop good intuition when working these types of problems.

“One should study mathematics simply because it helps to arrange one’s ideas” - M. W. Lomonossow

Antiderivatives of other trigonometric functions

Integrals involving products of \sin (x) and \cos (x)

Another strategy

Working with products of powers of \sec (x) and \tan (x)

Final thoughts

Integrals involving products of $\sin (x)$ and $\cos (x)$

Working with products of powers of $\sec (x)$ and $\tan (x)$