Trigonometric Integrals (Level 1) In the following series of videos we are going
to study techniques for evaluating integrals that contain certain combinations of trigonometric
functions, specifically those of the form sine of x raised to the power of m times cosine
of x raised to the power of n dx and secant of x raised to the power of m times tangent
of x raised to the power n dx, where both m and n are positive integers. To find the antiderivatives
of these types of integrals we need to use trigonometric identities to break the integrand
into combinations of trigonometric functions to which you can then apply the power rule
for integration by means of a u-substitution. Because we are going to rely heavily on trigonometric
identities, I recommend that you visit mathfortress.com and download a copy of the trigonometric identities
equation sheet located in the precalculus section. Alright we are first going to go
over trigonometric integrals involving powers of sine and cosine. Let’s begin by first
going over a trigonometric integral that we should already know how to do with the current
techniques at our disposal. Find the integral of sine of x raised to the
power 5 times cosine of x dx. The only techniques that we currently know
up to this point are u-substitution and integration by parts. Notice that we have a product of
two functions so you might be tempted to use integration by parts, but for the most part
you first want to check if u-substitution is a possible option. Taking a look at the
integrand we see that we can let u equal sine of x, and by taking the derivative of u with
respect to x we obtain cosine of x, notice that this function appears in the original
integral so we have a green light to carry on with the substitution step. Replacing sine
of x with u and du with cosine of x dx, we obtain the following integral in terms of
u. At this point it is just a matter of applying the power rule for integration, the integral
of u to the power of 5 du is equal to 1/6 times u to the power of 6 + C,
and the final step is to replace u with sine of x as follows. Doing that we obtain the final answer which
is equal to 1/6 times sin of x raised to the power of 6 plus C. This integral was relatively
easy to find because of the presence of a single cosine expression that was vital for
the u-substitution step to work. Keep this little detail in mind as we go over more challenging
trigonometric integrals. Alright with u-substitution fresh in our minds let’s
go ahead and try the next example.
Find the integral of cosine of x cubed dx. Ok notice that we can’t make a substitution
since we don’t have a single cosine of x factor, we actually have cosine of x cubed.
It seems that there is now way to solve the integral with our current techniques. This
is where we introduce a new technique so we can deal with these types of trigonometric
integrals. Notice that the cosine function has an odd power in this case its raised to
the power of 3, recall that we need a single cosine factor so that our u-substitution step
works out. The way we obtain a single cosine factor is by breaking apart the expression
into two distinct factors as follows, one factor is squared and other one is raised
to the power of 1. So now we have that single cosine factor we desperately needed. Now we
need an expression in terms of sine so that we can go ahead and use u substitution by
letting u = sin x, just the way we solved the first example, is there a way to rewrite
cosine squared in terms of sine? The answer is yes, this is where trigonometric identities
come into play, we can convert the remaining cosine squared factor to an expression involving
sine using the Pythagorean identity sine squared plus cosine squared = 1, by solving for cosine
squared within the identity itself, doing that we have that cosine squared is equivalent
to 1 minus sine squared. So now we can go back to our integral and substitute cosine
squared with this equivalent expression. We now have an expression in terms of sine and
the single cosine factor that we need to carry out the u-substitution step, so we go ahead
and let u = sin of x, then we find the derivative of u with respect to x, then we solve for
cosine of x dx and apply the substitutions throughout. Doing that we have the following
expression. Then it’s just a matter of finding the integral by using the power rule for integration,
the integral of 1 is just equal to u and the integral of u squared is equal to 1/3 times
u cubed and we add the constant C, the last step is to replace all the expressions that
contain the variable u with sine of x, doing that we obtain the final answer equal to sine
of x minus 1/3 times sine of x cubed plus C. So it turns out that the way we solve these
types of trigonometric integrals is by making use of trigonometric identities and u-substitution.
Alright let's go over the next example.
Find the integral of sine of x raised to the power of 6 times cosine of x raised to the power of 3 dx.
Alright once again notice that the cosine factor is raised to the power of 3 it’s
an odd number, in addition we also have a sine of x term in this case we have sine raised
to the power of 6 which is an even power. We have our expression in terms of sine, all
we need is a single cosine factor, like in the previous example let's go ahead and break
apart cosine cubed into cosine squared times cosine this way we obtain that single cosine
factor that is needed for the u-substitution step. Now that we have our single cosine factor
let’s go ahead and rewrite cosine squared by using the Pythagorean identity, this way
we have an expression in terms of sine only, doing that we have the following. Now that
we have our single cosine factor and the rest of the expression is in terms of sine only
lets proceed with the u-substitution step, so we let u = sin of x and take the derivative
of u with respect to x then we solve for cosine x dx, since this expression appears in the
original integral, then it’s just a matter of substituting the expressions though out,
doing that we obtain the following, then we go ahead and distribute u to the power of
6, next we use the power rule for integration to find the given integral, doing that we
obtain the following expression, the last step is to substitute the variables that contain
u with sine of x doing that we obtain the final answer equal to 1/7 times sine of x raised
to the power of 7 minus 1/9 times sine of x raised to the power of 9 plus C.
Notice that in this example the power of sine was even and positive in addition, the power
of cosine was odd and positive, in general we try to write an integrand involving powers
of sine that are even and powers of cosine that are odd in a form where we have only
one cosine factor (and the remainder of the integrand in terms of sine). With the help
of the Pythagorean identity sine squared plus cosine squared = 1 we can convert back and
forth between even powers of cosine and sine. For this case we want to use cosine squared
= 1 – sine squared. This way we can rewrite even powers of cosine into even powers of
sine. After that we let u = sin x and carry on with the u-substitution step and finally
we solve the integral in terms of u by first expanding the expression if needed, and applying
the power rule for integration. This is the first of 4 cases that we will cover though
out the video series. The one described here is when the power of sine is even and the
power of cosine is odd. Alright let's end the video by going over the final example.
Find the integral of sine of x raised to the power of 4 times cosine of x raised to the
power of 5 dx. We notice that this integral contains powers
of sine and cosine, we also see that sine has an even power and cosine has an odd power,
we always want to deal with the function that has an odd power, in this case we are going to break apart the
cosine expression into a single factor and a factor raised to the power of 4 as follows.
The next step is to rewrite cosine raised to the power of 4 as cosine squared raised
to the power of 2, this way we can substitute the cosine squared term with 1 - sine squared
as follows. Having isolated a single cosine factor and converted the remaining cosine
factors into expression in term of sine, we go ahead and carry on the substitution step
by letting u = sin of x, then we take the derivative of u with respect to x and solve
for cos of x dx, next we go ahead and substitute these expressions throughout.
Doing that we obtain the following integral,
next let's go ahead and expand the binomial, followed by distributing
u raised to the power of 4 doing that we obtain the following expression. Next let's go ahead
and use the power rule for integration to find the integral of the given expressions
Lastly let's go ahead and replace the variable u with
sine of x and as follows. Doing that we obtain the final answer equal to 1/5 times sine of
x raised to the power of 5 minus 2/7 times sine of x raised to the power of 7 plus 1/9
times sine of x raised to the power of 9 plus C.
Alright in our next video on trigonometric integrals we will go over the second case
when the power of sine is odd and the power of cosine is even.