Trigonometric Integrals (Level 2) In the previous video we started 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. Recall that we covered the case
when the power of cosine was odd. In this video we will go over an additional case.
Let’s go over the case when the power of sine is odd. Like in the previous video let’s
start with a couple of examples to illustrate the various steps that are key in solving
these types of trigonometric integrals. Find the integral of cosine squared of x times
sine of x dx. Like in the previous video always check to
see if u-substitution is a viable option before attempting other integration techniques since
u-substitution tends to be the easiest technique to execute. Scanning the integrand we see
that we can actually let u = cos x, making this substitution and taking the derivative
of u with respect to x, we obtain negative sine of x, then we go ahead and solve for sine
of x dx doing that we obtain the following, next let's go ahead and carry on the substitutions
doing that we obtain the following integral in terms of the variable u, next let's go ahead
and factor out the negative sign and proceed with the integration step, we find the integral
by applying the power rule for integration, the integral of u squared is equal to one
third times the u cubed and we add the constant C, the final step is to substitute the variable
u with the original expression in this case the original expression was cosine of x doing
that we obtain the following, and finally we go ahead and simplify the expression obtaining,
negative one third times cosine cubed plus C. Notice that in this integral we had an
expression where the power of cosine was even and the power of sine was odd. The reason
why the substitution step worked in this integral is because we had a single sine factor that
was vital for the substitution step to work. Make sure you place close attention to this
little detail in the proceeding examples. Alright let’s go ahead and try the next example.
Find the integral of sine cubed of 5x dx. Alright notice that we have a sine function
raised to power of 3 an odd integer, like in the previous example let’s try to use
u-substitution first, but before we can actually carry on a substitution we need to isolate
a single sin of (5x) factor so that the u-substitution step can work out. So let’s go ahead and
rewrite the integrand by breaking it apart into the a factor raised to the power of 2
and a factor raised to the power of 1, as follows, the reason why we want to break apart
an odd power function into an even power factor and a single factor is so that we can rewrite
the even power sine function into an expression in terms of cosine. We accomplish this by
using the Pythagorean identity sine squared plus cosine squared equals 1, Notice that
as long as both functions sine and cosine contain the same expression inside the argument of
the function this relation will hold, so we can actually use this version of the Pythagorean
identity to accommodate the problem at hand. Remember as long as both sine and cosine have
the same expression inside the argument this identity will hold. Using this version of
the Pythagorean identity and solving for sine squared of 5x, we obtain 1 minus cosine squared
of 5x, using this expression we go ahead and replace sine squared of 5x, as follows,
now that we have a single sine factor and the rest of the integrand in terms of cosine let's
go ahead and carry on the u-substitution step. So we let u = cosine of 5x, and we also find
the derivative of u with respect to x, make sure you apply the chain rule for this one,
doing that we have the following, next, let’s go ahead and solve for sine of x, after that
we go ahead and apply the substitutions throughout, doing that we have the following expression,
next, let’s go ahead and factor out the constant -1/5, followed by taking the integral
of the expression. The integral of 1 is equal to u, and the integral of u squared is equal
to 1/3 times u cubed, then we add the constant C. Next let's go ahead and simplify
the expression by distributing the constant negative 1/5. The final step is to substitute
the variables u with the original function in this case with cosine of 5x, doing that
we obtain the final answer equal to negative 1/5 times cosine of 5x plus 1/15 times cosine cubed
of 5x plus C. Notice that in this example the power on sine was odd, and we proceeded
by breaking it apart into a factor with an even power and a factor with a power of 1. Alright
Let’s try a slightly more challenging example.
Find the integral of cosine raised to the power of 4 times sine cubed dx
Alright similar to the last example notice
that we have powers of sine and cosine. The power on cosine is even and the power on sine
is odd. Recall that we want to break apart the odd power into an even power and a single
power function as follows. Having broken apart the odd powered sine function we go ahead
and replace sine squared with an equivalent expression in terms of cosine, so we go ahead
and use the Pythagorean identity sine squared plus cosine squared equals 1, and solve for
sine squared as follows, then we go ahead and replace sine squared with 1 minus cosine
squared. Now that we a single sine factor and the rest of the integrand is expressed in
terms of cosine, we go ahead and carry on the substitution step, so we let u = cosine of
x then we find the derivative of u with respective to x. Next we go ahead and solve for sine
of x dx since this expression occurs in the original integral. Then it’s just a matter
of substituting these expressions though out, doing that we obtain the following expression.
Factoring out the negative and distributing u to the power of 4 we obtain the following expression.
Now it’s just a matter of finding the integral by using the power rule for integration,
the integral of u to the power of 4 is just equal to 1/5 times u to the power of 5,
and the integral of u to the power of 6 is equal to 1/7 times u to the power of 7. Like always
we include the constant C, next we go ahead and distribute the negative sign, doing that
we obtain the following. The final step is to substitute back the original function
for u, in this case u was equal to cosine of x, doing that we obtain the final answer equal to
negative 1/5 times cosine raised to the power of 5 plus 1/7 times cosine
raised to the power of 7 plus C.
Notice that in this example the power on cosine was even and positive in addition, the power
of sine was odd and positive, in general we try to write an integrand involving powers
of cosine that are even and powers of sine that are odd in a form where we only have
one sine factor (and the remainder of the integrand in terms of cosine). With the help
of the Pythagorean identity sine squared plus cosine squared = 1 we can convert back and
forth between even powers of sine and cosine. For this case we want to use sine squared
= 1 – cosine squared. This way we can rewrite even powers of sine into even powers of cosine.
After that we let u = cosine of x and carry on with the u-substitution step and finally we
solve the integral by first expanding the expression if needed, and applying
the power rule for integration. This is the second of 4 cases that we will continue to
cover though out this video series. The one described here is when the power on cosine
is even and the power of sine is odd. Alright let’s end the video by going over the final example.
Find the integral of cosine squared times sine raised to the power of 5 dx.
Notice that we have a product involving powers of sine and cosine. So we need to identify
which function contains an odd power. Since this is what determines how we solve this
integral. In this case sine of x is raised to the power of 5 an odd number. So let’s go
ahead and break apart this function into an even power function and a single factor of
sine. Doing that we have the following. The next step is to rewrite sine raised to the
power of 4 as sine squared raised to the power of 2, this way we can substitute the sine
squared term with the Pythagorean identity 1 - cosine squared as follows. Now that we have
a single sine factor and the rest of the integrand written in terms of cosine we go ahead and
proceed with the u-substitution step so we let u = cosine of x, then we take the derivative
of u with respect to x. Then we solve for sin of x dx. After that we go ahead and substitute
these expressions though out, doing that we obtain the following. Next let’s go ahead
and factor out the negative sign and expand the binomial doing that we obtain the following
integral. Having expanded the binomial we go ahead and distribute u squared. Next we
go ahead and find the integral of each term by using the power rule for integration taking
the integral of each term yields the following expression. Don’t forget to add the constant
term when you are done taking the integral. Next let’s go ahead and distribute the negative
sign. The final step is to substitute back the original function for u, so we replace
the variable u with cosine of x. doing that we obtain the final answer equal to negative
1/3 times cosine cubed plus 2/5 times cosine raised to the power of 5 minus
1/7 times cosine raised to the power of 7 plus C.
Alright in our next video on trigonometric integrals we will go over the third case when
the power on sine and cosine are odd.