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Integration of Trigonometric Products and Powers

April 5th, 2009 |

Author: Merrin

In this section, we explore the next most complicated integration of trigonometric functions. The integrals of a product of sines and cosines to any integer powers in principle can be solved analytically.

$\displaystyle \int \text{sin}\,\, ^{n} x\text{cos}\,\, ^{m} x dx\quad \quad n,m\in {\mathbb Z}$

We can build up our table of integrals with these. It will be revealed later how other integrals transform into the ones we will solve here. Evaluating these integrals also allows us to practice our trigonometric identities and test our techniques of integration.

The problem can be divided into different classes. The usual classifications are

$\displaystyle \int \text{sin}\,\, ^{n} x\text{cos}\,\, ^{m} x dx\quad \quad n,m\in {\mathbb N}$

Some of the other cases with negative powers can be expressed as

$\displaystyle \int \text{tan}\,\, ^{n} x\text{sec}\,\, ^{m} x \quad \quad or\quad \, \displaystyle \int \text{cot}\,\, ^{n} x\text{csc}\,\, ^{m} x \quad n,m\in {\mathbb N}$

But these are not quite all the cases, don’t forget the following integrals.

$\displaystyle \int \cfrac{\text{sin}\,\,^nx}{\text{cos}\,\,^{n-m}x}dx \qquad \displaystyle \int \cfrac{\text{cos}\,\,^nx}{\text{sin}\,\,^{n-m}x}dx$

For the canonical first integral of products of sines and cosines to powers there are three cases that have different solutions: one where n is odd, one where m is odd, and one where n and m are both even. Having an odd power allows us to factor one of the terms and use it in a u substitution. When both powers are even we have to go to great lengths to evaluate the integrals by trigonometric identities.

Example 1. Evaluate the following

$\displaystyle \int \text{cos}\,\, ^{7} x\text{sin}\,\, ^{4}dx$

Solution 1. We notice that the power of cosine is odd so we can factor one of them out and do a u substitution.

$\cos^7 x \sin^4 xdx = \int \cos^6 x \sin^4 xd(\sin x)= \int (1-u^{2})^{3} u^4 du$
$\displaystyle \int (1-3u^{2}+3u^4 -u^6 )u^4 du=\cfrac{1}{5} u^5 -\cfrac{3}{7} u^7+\cfrac{3}{9} u^9 -\cfrac{1}{11} u^{11}+C$
${\int \text{cos}\,\, ^{7} x\text{sin}\,\, ^{4} xdx =\cfrac{1}{5} \text{sin}\,\, ^{5} x-\cfrac{3}{7} \text{sin}\,\, ^{7} x+\cfrac{1}{3} \text{sin}\,\, ^{9} x-\cfrac{1}{11} \text{sin}\,\, ^{11} x+C}$

Example 2. Evaluate the following

$\displaystyle \int \text{sin}\,\, ^{3} x \text{cos}\,\, ^{4} xdx$

Solution 2. We notice that the power of sine is odd so we can factor one out and do a u substitution.

$\int \text{sin}\,\, ^{3} x \text{cos}\,\, ^{4} xdx=\displaystyle \int \text{sin}\,\, ^{2} x\text{cos}\,\, ^{4} xd(-\text{cos}\,\, x)=-\displaystyle \int (1-u^{2} )u^{4} du }\\ {-\displaystyle \int (u^{4} -u^{6} )du =-\cfrac{1}{5} u^{5} +\cfrac{1}{7} u^{7} +C} \\ {\displaystyle \int \text{sin}\,\, ^{3} x \text{cos}\,\, ^{4} xdx=-\cfrac{1}{5} \text{cos}\,\, ^{5} x+\cfrac{1}{7} \text{cos}\,\, ^{7} x+C$

Example 3. Evaluate the following integral by using trigonometric identities

$\displaystyle \int \text{sin}\,\, ^{2} x\text{cos}\,\, ^{2} xdx$

Solution 3. The trigonometric identities come into play when the powers are even. The double angle formulas are necessary. Recall that

$\text{sin}\,\, 2x=2\text{sin}\,\, x\text{cos}\,\, x$

So we can rewrite the integrand as

$\text{sin}\,\, ^{2} x\text{cos}\,\, ^{2} x=\cfrac{1}{4} \text{sin}\,\, ^{2} 2x ^{}$

Now we use the double angle formula for cosine and sine and we have

$\text{sin}\,\, ^{2} x=\cfrac{1}{2} (1-\text{cos}\,\, 2x)$

So the integrand becomes

$sin ^2 x\cos^4 x=\cfrac{1}{8} (1-\cos 4x)$

The integral we compute is

$\displaystyle \int \text{sin}\,\,^2x\text{cos}\,\,^2x dx=\displaystyle \int \frac{1}{8}(1-\text{cos}\,\, 4x)dx = \cfrac{x}{8}-\cfrac{\text{sin}\,\,4x }{32}+C$

Life is hard when both powers are both even.

Other trigonometric identities can come into play in more complicated integrals. These identities are used to split products of sines and cosines into fundamental integrals. For example,

$\text{sin}\,\, A\text{sin}\,\, B=\cfrac{1}{2} \left[\text{cos}\,\, (A-B)-\text{cos}\,\, (A+B)\right]$

The next sets of integrals we will consider are of the form

$\displaystyle \int \text{tan}\,\, ^{n} x\text{sec}\,\, ^{m} xdx \quad \quad \, n,m\in {\mathbb N}$

This integral is similar to the one we did before except that we put the cosines in the denominator and the sines in the numerator. Before we break up the integral into classes we should recall some basic trigonometric and derivative identities.

$\cfrac{d}{dx} \text{tan}\,\, x=\text{sec}\,\, ^{2} x\qquad 1+\text{tan}\,\, ^{2} x=\text{sec}\,\, ^{2} x \qquad \cfrac{d}{dx} \text{sec}\,\, x=\text{sec}\,\, x\text{tan}\,\, x$

Example 4. Evaluate the following integral

$\displaystyle \int \text{tan}\,\, ^{n} x\text{sec}\,\, ^{2} xdx$

Solution 4. Since we notice that the square of secant is the derivative of tangent we find the answer as

$\displaystyle \int \text{tan}\,\, ^{n} x\text{sec}\,\, ^{2} xdx =\displaystyle \int u^{n} du =\cfrac{1}{n+1} u^{n+1} +C=\cfrac{1}{n+1} \text{tan}\,\, ^{n+1} x$

Similarly, in any other integral that has an even power of secant two are
factorable and reduce the others by the trigonometric identity. This is the form it looks like

${\displaystyle \int \text{tan}\,\, ^{n} x\text{sec}\,\, ^{2m} xdx } = {\displaystyle \int \text{tan}\,\, ^{n} x(1+\text{tan}\,\, ^{2} x)^{2m-2} \text{sec}\,\, ^{2} xdx }\\= \displaystyle \int u^n(1+u^2)^{2m-2}du$

Example 5. Evaluate the following integral, where the power of secant is odd

$\displaystyle \int \text{tan}\,\, ^{3} x\text{sec}\,\, xdx$

Solution 5. Since the power of secant is odd we use the other derivative identity which is

$\cfrac{d}{dx} \text{sec}\,\, x=\text{sec}\,\, x\text{tan}\,\, x$

So our integral becomes

$\displaystyle \int \text{tan}\,\, ^{3} x\text{sec}\,\, xdx =\displaystyle \int (\text{sec}\,\, ^{2} x-1)d(\text{sec}\,\, x)=\cfrac{1}{3} \text{sec}\,\, ^{3} x-\text{sec}\,\, x+C$

As long as the original power of tangent is odd along with the power of secant being odd then just evaluate the integral right away using the derivative
of secant.

Example 6. Evaluate the integral

$\displaystyle \int \text{tan}\,\,^4x \text{sec}\,\,^xdx$

Solution 6. What happens if the power of secant is odd and the power of tangent is even? Just convert everything into secant and integrate powers of secant by the reduction formula.

$\displaystyle \int \text{tan}\,\, ^{4} x\text{sec}\,\, xdx=\displaystyle \int (\text{sec}\,\, ^{2} -1)^{2} \text{sec}\,\, xdx=\displaystyle \int (\text{sec}\,\, ^{5} x-2\text{sec}\,\, ^{3} x+\text{sec}\,\, x)dx$

We have considered all the cases when secant is even. In these cases, the square of secant is factored out and used to integrate the tangents. When tangent and secant are both odd then a secant times a tangent is factored and the remaining tangents are converted into secants. When tangent is even and secant is odd, all the powers of tangent are converted to secant and each power of secant is integrated by the reduction formula.

Moving onto the next case we have the roles of sine and cosine reversed.

$\displaystyle \int \text{cot}\,\, ^{n} x\text{csc}\,\, ^{m} xdx \quad \quad \quad \quad n,m\in {\mathbb N}$

We won’t discuss these solutions because they are too similar to what we did before. Later we will learn a transformation that can in reverses all of the sines and cosines in an integral equation getting the negative signs correct. For now we take it that these integrals are solved by the
same methods.

We have a solution method for the following cases

$\displaystyle \int \text{sin}\,\,^n x \text{cos}\,\,^m x dx \qquad \displaystyle \int \cfrac{\text{sin}\,\,^nx}{\text{cos}\,\,^{n+m}x}dx \qquad\displaystyle \int \cfrac{\text{cos}\,\,^nx}{\text{sin}\,\,^{n+m}x}dx$

The cases we are missing are

$\displaystyle \int \cfrac{\text{sin}\,\,^nx}{\text{cos}\,\,^{n-m}x}dx \qquad\displaystyle \int \cfrac{\text{cos}\,\,^nx}{\text{sin}\,\,^{n-m}x}dx$

Let us try some examples and see what we can say about these types of integrals.

Example 7. Find the integral of

$\displaystyle \int \cfrac{\text{sin}\,\, ^{5} x}{\text{cos}\,\, x} dx$

Solution 7. As we discussed earlier this integral doesn’t fit one of the three forms. What we can do is to write this as

$\displaystyle {\int \cfrac{\sin ^{5} x}{\cos x} dx }={-\int \cfrac{(1-\cos ^{2} x)^{2} d(\cos x)}{\cos x} =-\int \cfrac{(1-u^{2} )^{2} }{u} du} \\ = {-\int \cfrac{(1-2u^{2} +u^{4} )}{u} du=\int (-u^{-1} +2u-u^{3} )du }$

This worked out nicely because we used our old trick of an odd power of sine with u substitution. If we had pumped up the power of cosine a
bit then we would just use the reduction formula for powers of secants.

Example 8. Let us tweak the last integral a little bit to make it harder, find

$\displaystyle \int \cfrac{\text{sin}\,\, ^{4} x}{\text{cos}\,\, x} dx$

Solution 8. We are going to have to be a little cleverer with this one.

$\displaystyle \int \cfrac{\text{sin}\,\, ^{4} x}{\text{cos}\,\, x} dx =\displaystyle \int \cfrac{\text{sin}\,\, ^{4} x}{\text{cos}\,\, ^{2} x} \text{cos}\,\, xdx =\displaystyle \int \cfrac{\text{sin}\,\, ^{4} x}{1-\text{sin}\,\, ^{2} x} d(\text{sin}\,\, x)=\displaystyle \int \cfrac{u^{4} }{1-u^{2} } du$

Well now we go back into partial fractions mode. After polynomial long division and solving for the coefficients of the remainder in the partial fractions expansion we have

$\cfrac{u^{4} }{1-u^{2} } =-u^{2} -1-\cfrac{1}{2(u-1)} +\cfrac{1}{2(u+1)}$

This can be easily integrated to be

$\int \cfrac{u^{4} }{1-u^{2} } du } = {-\cfrac{1}{3} u^{3} -u-\cfrac{1}{2} \ln \left|u-1\right|+\cfrac{1}{2} \ln \left|u+1\right|+C} \\ {u} = {\text{sin}\,\, x$

For these last types of integrals, if the odd power trick doesn’t lead anywhere, then try devising a substitution to evaluate the integral as a rational function by partial fractions. Even is we had bumped up the power of the denominator it would have still be solvable by partial fractions. The denominator is always entirely factored so partial fractions is just a matter of algebra and will always succeed.

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