# 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.

(1+cos(4x))/2 I simply distributed the negative sign and 1/2 to 1 and cos(4x). Its like distributing 1/2*(1+x) = 1/2 +x/2

I will keep that in mind as I create the next videos. Let every one know about Math Fortress.

you're really good! thanks a lot.

Thank you! Let every one know how useful these videos are.

hey can anyone help me with this problem I'm stuck on it goes

integral(sin^5xcos^-2x)

the answer in the book goes like

secx+2cosx-(cos^3x)/x

the number after the carrot is the exponent NOT including the x

Amazing way of teaching.

I have no idea if you even use youtube anymore, but the last problem, can: Sin^3(2x)/48 + x/8 – x/16 – Sin(4x)/64 be a correct answer as well? I notice you just created a common denominator to add the two x terms to get x, but is that neccessary? Thanks! Also great videos!

what if both powers are 4?