8 Techniques of Integration

8.2 Trigonometric Integrals

Trigonometric functions are useful for describing periodic behavior. This section describes several techniques for finding antiderivatives of certain combinations of trigonometric functions.

Integrals of the form sinmxcosnxdx

In learning the technique of Substitution, we saw the integral sinxcosxdx in Example 5.5.4. The integration was not difficult, and one could easily evaluate the indefinite integral by letting u=sinx or by letting u=cosx. This integral is easy since the power of both sine and cosine is 1.

We generalize this and consider integrals of the form sinmxcosnxdx, where m,n are nonnegative integers. Our strategy for evaluating these integrals is to use the identity cos2x+sin2x=1 to convert high powers of one trigonometric function into the other, leaving a single sine or cosine term in the integrand. We summarize the general technique in the following Key Idea.

Key Idea 8.2.1 Integrals Involving Powers of Sine and Cosine

Consider sinmxcosnxdx, where m,n are nonnegative integers.

  1. (a)

    If m is odd, then m=2k+1 for some integer k. Rewrite




    where u=cosx and du=-sinxdx.

  2. (b)

    If n is odd, then using substitutions similar to that outlined above we have


    where u=sinx and du=cosxdx.

  3. (c)

    If both m and n are even, use the half-angle identities


    to reduce the degree of the integrand. Expand the result and apply the principles of this Key Idea again.

We practice applying Key Idea 8.2.1 in the next examples.

Example 8.2.1 Integrating powers of sine and cosine

Evaluate sin5xcos8xdx.

SolutionThe power of the sine factor is odd, so we rewrite sin5x as


Our integral is now (1-cos2x)2cos8xsinxdx. Let u=cosx, hence du=-sinxdx. Making the substitution and expanding the integrand gives

(1-cos2x)2cos8xsinxdx =-(1-u2)2u8du
Example 8.2.2 Integrating powers of sine and cosine

Evaluate sin5xcos9xdx.

SolutionBecause the powers of both the sine and cosine factors are odd, we can apply the techniques of Key Idea 8.2.1 to either power. We choose to work with the power of the sine factor since that has a smaller exponent.

We rewrite sin5x as

sin5x =sin4xsinx

This lets us rewrite the integral as


Substituting and integrating with u=cosx and du=-sinxdx, we have

(1-2cos2x+cos4x) sinxcos9xdx

Instead, another approach would be to rewrite cos9x as

cos9x =cos8xcosx

We rewrite the integral as


Now substitute and integrate, using u=sinx and du=cosxdx.


Technology Note:

The work we are doing here can be a bit tedious, but the skills developed (problem solving, algebraic manipulation, etc.) are important. Nowadays problems of this sort are often solved using a computer algebra system. The powerful program Mathematica® integrates sin5xcos9xdx as


which clearly has a different form than our second answer in Example 8.2.2, which is margin: g(x)f(x)123-0.0020.0020.004xy Figure 8.2.1: A plot of f(x) and g(x) from Example 8.2.2 and the Technology Note. Λ


Figure 8.2.1 shows a graph of f and g; they are clearly not equal, but they differ only by a constant: g(x)=f(x)+C for some constant C. So we have two different antiderivatives of the same function, meaning both answers are correct.

Example 8.2.3 Integrating powers of sine and cosine

Evaluate sin2xdx.

SolutionThe power of sine is even so we employ a half-angle identity, algebra and a u- substitution as follows:

sin2xdx =1-cos(2x)2dx
Example 8.2.4 Integrating powers of sine and cosine

Evaluate cos4xsin2xdx.

SolutionThe powers of sine and cosine are both even, so we employ the half-angle formulas and algebra as follows.

cos4xsin2xdx =(1+cos(2x)2)2(1-cos(2x)2)dx

The cos(2x) term is easy to integrate. The cos2(2x) term is another trigonometric integral with an even power, requiring the half-angle formula again. The cos3(2x) term is a cosine function with an odd power, requiring a substitution as done before. We integrate each in turn below.


Finally, we rewrite cos3(2x) as


Letting u=sin(2x), we have du=2cos(2x)dx, hence

cos3(2x)dx =(1-sin2(2x))cos(2x)dx

Putting all the pieces together, we have


The process above was a bit long and tedious, but being able to work a problem such as this from start to finish is important.

Integrals of the form tanmxsecnxdx

When evaluating integrals of the form sinmxcosnxdx, the Pythagorean Theorem allowed us to convert even powers of sine into even powers of cosine, and vice versa. If, for instance, the power of sine was odd, we pulled out one sinx and converted the remaining even power of sinx into a function using powers of cosx, leading to an easy substitution.

The same basic strategy applies to integrals of the form tanmxsecnxdx, albeit a bit more nuanced. The following three facts will prove useful:

  • ddx(tanx)=sec2x,

  • ddx(secx)=secxtanx , and

  • 1+tan2x=sec2x (the Pythagorean Theorem).

If the integrand can be manipulated to separate a sec2x term with the remaining secant power even, or if a secxtanx term can be separated with the remaining tanx power even, the Pythagorean Theorem can be employed, leading to a simple substitution. This strategy is outlined in the following Key Idea.

Key Idea 8.2.2 Integrals Involving Powers of Tangent and Secant

Consider tanmxsecnxdx, where m and n are nonnegative integers.

  1. (a)

    If n is even, then n=2k for some integer k. Rewrite secnx as




    where u=tanx and du=sec2xdx.

  2. (b)

    If m is odd and n>0, then m=2k+1 for some integer k. Rewrite tanmxsecnx as




    where u=secx and du=secxtanxdx.

  3. (c)

    If n is odd and m is even, then m=2k for some integer k. Convert tanmx to (sec2x-1)k. Expand the new integrand and use Integration By Parts, with dv=sec2xdx.

  4. (d)

    If m is even and n=0, rewrite tanmx as



    tanmxdx=tanm-2sec2xdxapply rule #1-tanm-2xdxapply rule #4 again.

The techniques described in items 1 and 2 of Key Idea 8.2.2 are relatively straightforward, but the techniques in items 3 and 4 can be rather tedious. A few examples will help with these methods.

Example 8.2.5 Integrating powers of tangent and secant

Evaluate tan2xsec6xdx.

SolutionSince the power of secant is even, we use rule #1 from Key Idea 8.2.2 and pull out a sec2x in the integrand. We convert the remaining powers of secant into powers of tangent.

tan2xsec6xdx =tan2xsec4xsec2xdx
Now substitute, with u=tanx, with du=sec2xdx.
We leave the integration and subsequent substitution to the reader. The final answer is

We derived integrals for tangent and secant in Section 5.5 and will regularly use them when evaluating integrals of the form tanmxsecnxdx. As a reminder:

tanxdx =ln|secx|+C
secxdx =ln|secx+tanx|+C
Example 8.2.6 Integrating powers of tangent and secant

Evaluate sec3xdx.

SolutionWe apply rule #3 from Key Idea 8.2.2 as the power of secant is odd and the power of tangent is even (0 is an even number). We use Integration by Parts; the rule suggests letting dv=sec2xdx, meaning that u=secx.
u=secxdv=sec2xdxdu=?v=?    u=secxdv=sec2xdxdu=secxtanxdxv=tanx Figure 8.2.2: Setting up Integration by Parts.

Employing Integration by Parts, we have

sec3xdx =secxusec2xdxdv
This new integral also requires applying rule #3 of Key Idea 8.2.2:
margin: Note: Remember that in Example 5.5.8, we found that secxdx=ln|secx+tanx|+C Λ

In previous applications of Integration by Parts, we have seen where the original integral has reappeared in our work. We resolve this by adding sec3xdx to both sides, giving:

2sec3xdx =secxtanx+ln|secx+tanx|
sec3xdx =12(secxtanx+ln|secx+tanx|)+C.

We give one more example.

Example 8.2.7 Integrating powers of tangent and secant

Evaluate tan6xdx.

SolutionWe employ rule #4 of Key Idea 8.2.2.

tan6xdx =tan4xtan2xdx
We integrate the first integral with substitution, u=tanx and du=sec2xdx; and the second by employing rule #4 again.
Again, use substitution for the first integral and rule #4 for the second.

Integrals of the form cotmxcscnxdx

Not surprisingly, evaluating integrals of the form cotmxcscnxdx is similar to evaluating tanmxsecnxdx. The guidelines from Key Idea 8.2.2 and the following three facts will be useful:

ddx(cotx) =-csc2x
ddx(cscx) =-cscxcotx,and
csc2x =cot2x+1
Example 8.2.8 Integrating powers of cotangent and cosecant

Evaluate cot2xcsc4xdx

SolutionSince the power of cosecant is even we will let u=cotx and save a csc2x for the resulting du=-csc2xdx.

cot2xcsc4xdx =cot2xcsc2xcsc2xdx

The integration and substitution required to finish this example are similar to that of previous examples in this section. The result is


Integrals of the form sin(mx)sin(nx)dx, cos(mx)cos(nx)dx, and sin(mx)cos(nx)dx.

Functions that contain products of sines and cosines of differing periods are important in many applications including the analysis of sound waves. Integrals of the form


are best approached by first applying the Product to Sum Formulas of Trigonometry found in the back cover of this text, namely

sin(mx)sin(nx) =12[cos((m-n)x)-cos((m+n)x)]
cos(mx)cos(nx) =12[cos((m-n)x)+cos((m+n)x)]
sin(mx)cos(nx) =12[sin((m-n)x)+sin((m+n)x)]
Example 8.2.9 Integrating products of sin(mx) and cos(nx)

Evaluate sin(5x)cos(2x)dx.

SolutionThe application of the formula and subsequent integration are straightforward:

sin(5x)cos(2x)dx =12[sin(3x)+sin(7x)]dx

Integrating other combinations of trigonometric functions

Combinations of trigonometric functions that we have not discussed in this chapter are evaluated by applying algebra, trigonometric identities and other integration strategies to create an equivalent integrand that we can evaluate. To evaluate “crazy” combinations, those not readily manipulated into a familiar form, one should use integral tables. A table of “common crazy” combinations can be found at the end of this text.

These latter examples were admittedly long, with repeated applications of the same rule. Try to not be overwhelmed by the length of the problem, but rather admire how robust this solution method is. A trigonometric function of a high power can be systematically reduced to trigonometric functions of lower powers until all antiderivatives can be computed.

The next section introduces an integration technique known as Trigonometric Substitution, a clever combination of Substitution and the Pythagorean Theorem.

Exercises 8.2


Terms and Concepts

  1. 1.

    T/F: sin2xcos2xdx cannot be evaluated using the techniques described in this section since both powers of sinx and cosx are even.

  2. 2.

    T/F: sin3xcos3xdx cannot be evaluated using the techniques described in this section since both powers of sinx and cosx are odd.

  3. 3.

    T/F: This section addresses how to evaluate indefinite integrals such as sin5xtan3xdx.

  4. 4.

    T/F: Sometimes computer programs evaluate integrals involving trigonometric functions differently than one would using the techniques of this section. When this is the case, the techniques of this section have failed and one should only trust the answer given by the computer.


In Exercises 5–32., evaluate the indefinite integral.

  1. 5.


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In Exercises 33–40., evaluate the definite integral.

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  8. 40.


  1. 41.

    Find the area between the curves y=sin2x and y=cos2x on the interval [π/4,3π/4].

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