Trigonometric Integrals (Level 5) In the previous video we started solving trigonometric
integrals where the powers on sine and cosine were even. In this video we are going to go
over more challenging examples. Let’s jump straight into the first example.
Find the integral of sine squared of x times cosine squared of x dx.
This is an integral that contains powers of sine and cosine that are even, so
we need to make use of the half angle identities. So we first need to replace sine squared and
cosine squared with their respective half angle identity as follows. Recall that the
purpose of using these identities is to reduce the even power on sine and cosine into odd
powers of cosine. This way we are free to use the rest of our integration techniques
learned from the previous videos. Next we go ahead and factor out the constant obtained
from both half angle replacements, in this case we factor out two one half's, then we multiply
them to obtain one fourth. The next step is to multiply both binomials and collect like terms doing
that we obtain the following expression. Notice that we ended with another even powered
cosine term, this means that we have to use the half angle identity once more, so we do
just that and obtain the following expression. Simplifying the expression and collecting
like terms we obtain the following. Here, we go ahead and factor out the constant one half
and multiply it with the constant that we factored out in the previous steps obtaining
the constant 1/8. Now that we have an odd powered cosine term we proceed with the integration
step. So taking the integral term by term we have that the integral of 1 is equal to
x and the integral of cosine of 4x requires a u-substitution from this point on, I am
going to skip most u-substitution steps to save time. In any case I will make sure that
I mention when u-substitution is required to find the integral of an expression. So
using u-substitution we find that the integral of cosine of 4x is equal to sine of 4x over
4. The final step is to distribute the constant 1/8 doing that we obtain the final answer
equal to x over 8 minus sine of 4x over 32 plus C.
It turns out that there is often more than one way in finding integrals in which both
the exponents on sine and cosine are even. Let’s go ahead and solve the same problem
using a slightly different method. Since the integrand contains a product containing both sine
and cosine and the powers are essentially the same, they both contain a power of 2.
We can actually use the following double angle formula to solve the integral.
sine of 2x equals 2 times sine of x times cosine of x
Notice that we can solve for the product sine
times cosine as follows. Then we rewrite the integrand as a product in this case the quantity
sine times cosine squared so that we can replace this expression with the quantity ½ times
sine of 2x squared. Then we go ahead and square the expression, doing that we obtain the following.
Next we factor out the constant ¼ followed by a half angle identity replacement so we
replace sine squared of 2x with 1-cosine of 4x over 2 notice that we took into account
the argument of the expression we need to make sure we double the argument when using
a half angle identity. At this point the steps in finding the integral are similar
if not identical to the way we solved this problem the first time around. All that is
left to do is to take the integral term by term. The integral of 1 is just equal x and
the integral of cosine of 4x requires a u substitution which yields sine of 4x over 4.
Making sure we include the constant C. The final step is to distribute the constant 1/8
doing that we obtain the final answer equal to x over 8 minus sine of 4x over 32 plus C.
In this example we went over two different solution methods that gave the same answer.
Note that this will not always happen. In fact, more often than not we will obtain different
answers. This difference will tend to differ by no more than a constant. In this example
It just happened to occur that both answers ended being the same.
In general when we have products of sines and cosines in which both the exponents are even
we will need to use a series of half angle and/or double angle formulas to reduce the
integral into a form that we can integrate. For the most part using half angles identities
is usually the way to go. Also, the larger the exponents the more we’ll need to use
these formulas. Let’s go over an example that illustrates this repeated use of the formulas.
Find the integral of cosine raised to the power of 4 of x times sine squared of x dx.
Alright notice that we have powers of sine
and cosine, in addition both sine and cosine contain even powers this means that we need
to repeatedly make use of the half angle identities. Let’s start by rewriting cosine raised to
the power of 4 into cosine squared raised to the power of 2, this way we can replace
cosine squared with its corresponding half angle identity. The next step is to do just
that we replace cosine squared and sine squared with their respective half angle identities
as follows. We then square the numerator and denominator on the left expression. Then we
factor out the constants in this case we have ¼ for the left expression and ½ for the
right expression. Multiplying these constants we obtain 1/8, we also want to expand the
following binomial remember we need to use the FOIL method for this one. Expanding the
expression we obtain the following. Now we have to multiply the trinomial with the binomial,
distributing term by term we obtain the following expression. Next we collect like terms and
simplify the expression. From this point it’s just a matter of finding the integral term
by term. So the integral of 1 is just equal to x, and the integral of cosine of 2x requires
a u-substitution carrying out the u-substitution step results in the following expression.
Now the integral of the rest of the terms is not as straight forward, the integral
of cosine squared of 2x, contains an even power this means that we have to apply the
half angle identity once again to find this particular integral, so we go ahead and replace
cosine squared of 2x with 1+cosine of 4x over 2 once again make sure you double the argument
when carrying out the half angle replacement. Next, we factor out the constant ½ and proceed
with the integration step, the integral of 1 is equal to x, and the integral of cosine
of 4x can be found via a u-substitution. Doing that we obtain sine of 4x over 4, then we go
ahead and distribute the constant 1/2 . The expression we just found is the integral of
cosine squared of 2x. Now we need to find the integral of the last term cosine cubed
of 2x, notice that this expression contains an odd power. Recall that this is essentially a
case I trigonometric integral. So we need to decompose the integrand into an even powered
function and a single function of degree 1 as follows. Next we use the Pythagorean identity
to rewrite the even power of cosine into an even power in terms of sine. This
way we can go ahead and carry out a u-substitution using the following substitutions. Doing
that we obtain the following integral in terms of the variable u. Then it’s just a matter
of finding the integral term by term, the integral of 1 is equal to u, and the integral
of u squared is equal to 1/3 times u cubed, next we distribute the constant and substitute
back the original expression for u doing that we obtain the following integral. With this
final expression we essentially found our integral so we add the constant C. Now it’s
just a matter of cleaning this answer a bit, so we distribute the negative signs as follows,
then we collect like terms, two of the terms cancel out which is always nice to see. The
Final step is to distribute the constant 1/8 doing that we obtain the final answer equal
to x over 16 minus sine of 4x over 64 plus sine cubed of 2x over 48 plus C.
This problem was pretty epic not only did we make use of the half angle identities we
also had to use u-substitution and a Pythagorean identity. This is why
you should be comfortable applying all the integration techniques learned up to this point, so that
you can solve integrals that require multiple integration techniques. Now that we have covered
all four cases that involve powers of sine and cosine, we are ready to cover the case
where the integrand contains products of sines and cosines that contain different arguments.
This will be the topic of our next video.