Trigonometric Integrals (Level 4) In the previous video we continued solving
trigonometric integrals of the form sine of x raised to the power of m times cosine of
x raised to the power of n where both m and n are positive integers. Up to this point
we essentially covered three distinct cases. All of them required that at least one of
the trigonometric functions contained an odd exponent. In the preceding video we covered
the third case where both the powers on sine and cosine were odd. In this video we will
go ahead and cover the 4th and last case specifically the case when both the powers on sine and
cosine are even. Recall that in the previous videos an odd
power of sine or cosine enabled us to separate a single factor and convert the remaining
even power in terms of another trigonometric function by using the Pythagorean identity.
If the integrand contains even powers of both sine and cosine, this strategy inevitably
fails. The strategy that we want to use when we are face to face with these types of integrals
is to repeatedly use the following trigonometric identities sine squared of x equals
1 minus cosine of 2x over 2, and cosine squared of x equals 1 plus cosine of
2x over 2. These identities are modified versions of the half angle formulas that you encountered
in your trigonometry or precalculus class. The idea behind using these identities is
to rewrite the even powered functions, into odd power functions. It turns out that these
identities do just that, since they convert even powered functions into odd powered functions
of degree 1 these identities are also known as power reduction formulas. Then from here
we go ahead and proceed with the integration step. Let’s illustrate this technique by
fist going over a basic example. Find the integral of sine squared of x dx.
Alright notice that we have an integral that contains even powers of sine and cosine.
In this case we only have an even power of sine. But technically cosine also has an even power.
Recall that in this video series we are dealing with trigonometric integrals of
the following form. For this example the value for m is equal to 2 and the value of n is
equal to 0 which is technically an even number since it shares many of the properties that
even numbers posses. This means that we are dealing with a trigonometric integral where
both the powers on sine and cosine are even. Alright, if we attempt to solve this integral
by using the techniques from the previous video, specifically, using the Pythagorean
identity sin^2 + cosine^2 = 1 to replace sin^2 with an expression in terms of cosine.
We would obtain the following expression. Notice that we can easily take the integral of 1
this is just equal to x, but then this is where we encounter our first problem we now
need to take the integral of cosine squared, an even powered function, now what? Well we
can try and use the Pythagorean identity once more and rewrite cosine squared with an expression
in terms of sine as follows. Then if we try to take the integral of the resulting expression
we end up with the following, notice that the linear terms cancel out and we end up
with the original integral, all we did was to go around in a big circle.
This is why you don’t want to use the Pythagorean identity rather you want to use the modified
half angle identities also known as power reducing formulas. So let’s go ahead and
use the following identity and replace sine squared with 1 - cosine of 2x over 2. As follows.
Notice that we started with an expression of degree 2 an even power, and ended with an expression
of degree 1 an odd power. Now we are free to use the rest of our integration techniques
that we learned from the previous videos to find the integral, so let’s go ahead and
factor out the constant ½, followed by a term by term integration, so the integral
of 1 is just equal to x, and the integral of cosine of 2x requires a u-substitution,
so we let u equal 2x and then we find the derivative of u with respect to x then we
solve for dx. Next we carry out the substitutions throughout, followed by finding the integral
of the expression in terms of u. The last step is to substitute back the original expression
for u. Doing that we obtain the following. So the integral of cosine of 2x is equal to
½ times sine of 2x, then we make sure we include the constant C. The final step is
to distribute the constant one half, doing that we obtain the final answer equal to x
over 2 minus sine of 2x over 4 plus C. Alternatively we can also make use of the double angle identity
sin of 2x = 2 times sine of x times cosine of x to rewrite the sine of 2x expression
in terms of sine and cosine. Doing that we obtain an alternative answer equal to x over
2 minus sine of x times cosine of x over 2 plus C.
Alright lets practice some more and try the second example.
Find the integral of cosine squared of 3x dx.
Notice that this integral contains powers of sine and cosine that are even in this
case, the power on sine is 0, this is the reason why the sine term is nowhere to be
found in the integrand. But, this is not the case for cosine which contains an even power greater than 0.
This means that we need to make use of the modified half angle identities to reduce the
even power into an odd power. Before we carry out the half angle identity substitution,
Notice that the argument on cosine is something other than x in this case its 3x as a result,
we need to take into account this distinction in the argument when we use the half angle
identity. The way we compensate for this distinction is by observing that the argument
on cosine in the right side of the identity is twice the argument of cosine on the left
side of the identity. This means that when we carry out the half angle identity substitution
we need to make sure we double the argument for cosine. So we start by substituting cosine
squared of 3x with 1 plus cosine of 6x over 2. Notice that we double the argument from 3x to 6x
when we carried out the half angle substitution.
Next we factor out the constant ½
then we go ahead and find the integral term by term, the integral of 1 is equal to x and
the integral of cosine of 6x requires a u-substitution, so we let u = 6x and we also need to find
the derivative of u with respect to x after that we solve for dx, next we substitute the
expressions throughout doing that we obtain the following. Factoring out the 1/6 and finding
the integral of cosine of u we obtain the following. Then, we substitute back the original
expression for u. So the integral of cosine of 6x is equal to 1/6 times sine of 6x. Making
sure we add the constant C. After that, it’s just a matter of distributing the ½ doing
that we obtain the final answer equal to x over 2 plus sine of 6x over 12 plus C. Alternatively
we can use the double angle formula and write the answer in terms of sine and cosine.
Once again we need to take into account the argument of the trigonometric function. For this identity
the argument of the expression on the right side is half the argument of the expression
on the left side. So we can replace sine of 6x with 2 times sin of 3x times cosine of
3x. Simplifying the expression we obtain an alternative answer of the integral equal to
x over 2 plus sine of 3x times cosine of 3x over 6 plus C. Make sure you take into account
the argument of the trigonometric function when using identities you might need to take
the expressions inside the argument into account.
In general if both the powers on sine and cosine are even and nonnegative. We make repeated
use of the following half angle identities so that we can convert the integrand into
odd powers of cosine. Then we proceed by applying the same techniques that we used for case
I and case II from the previous videos. In addition to the half angle identities it is
sometimes helpful to use the double angle identity sin of 2x = 2(sin x)(cos x), especially
when we solve for the product sine times cosine. After all we are dealing with integrals that
contain products of sine and cosine. This is the 4th and last case that we will cover
in this video series of integrals that contain powers of sine and cosine that are greater
than or equal to zero. Alright now that we know how to use our new identities to solve
even powered sine and cosine functions we are ready to solve integrals that contain products
of sine and cosine that contain even powers. In our next video we will cover slightly more
complicated case 4 integrals.