Calculus II: Trigonometric Integrals (Level 5 of 7) | Even Power on Sine and Cosine II


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.

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