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## Integrating using trigonometric identities

Aktuel tid:0:00Samlet varighed:5:33

# Integral of sin^4(x)

## Video udskrift

- [Voiceover] Let's see if we can take the indefinite integral of
sine of x to the fourth dx. And like always, pause the video and see if you can work
through it on your own. So if we had an odd exponent up here, whether it was a sine or a cosine, then the technique I would use, so if this was sine to the third of x, I would separate one
of the sine of xs out, so I would rewrite it as sine
squared x times sine of x, and then I would convert this using the Pythagorean Identity, and then when I distribute the sine of x, I'd be able to use u-substitution. We've done that in
previous example videos. You could have done this if it was cosine to the third of x
as well, or to the fifth, or to the seventh, if
you had an odd exponent. But here we have an even exponent. So what do we do? So the technique we will use, and I guess you could call it a trick, the technique or trick to use is once again you want to
algebraically manipulate this so you can use integration techniques that we are familiar with. And in this case we would want to use the Double Angle Identity. The Double Angle Identity
tells us that sine squared of x is equal to 1/2 times one
minus cosine of two x. So how can I apply this over here? Our original integral is
just the same thing as, this is going to be the same thing as the integral of sine of x
squared, all of that squared, dx. Now I can make this substitution. So this is the same thing as this, which is of course the same thing as this by the Double Angle Identity. So I could rewrite it
as the integral of 1/2 times one minus cosine of two x, and then all of that squared dx. And what is that going to be equal to? That's going to be equal to, let's see. 1/2 squared is 1/4, so
I can take that out. So we get 1/4 times the integral of... I'm just going to square
all this business. One squared, which is one,
minus two cosine of two x plus cosine squared of two x dx. Fairly straightforward to
take the indefinite integral, or to take the antiderivative
of these two pieces. But what do I do here? Once again I've got an even exponent. Let's apply the Double
Angle Identity for cosine. We know that cosine squared of two x is going to be equal to 1/2 times one plus cosine of double this
angle, so cosine of four x. Once again just make the substitution. This is going to be equal to... Actually let me just write it this way. 1/4, and maybe I'll switch... I'll just keep it in this color. 1/4 times one minus two
cosine two x plus 1/2, I'll just distribute the 1/2. Plus 1/2 plus 1/2 cosine of four x dx. Let's see, I could take this
1/2 and add it to this one, and that's going to get me 3/2, so add those together,
I'm going to get 3/2. So let me rewrite this as, I'm
in the home stretch really, this is going to be equal to
1/4 times the integral of 3/2 minus two cosine of two x. The derivative of four x is four, so it'd be great if I had a four out here. Let me rewrite 1/2 as four over eight. Really I'm just multiplying
and dividing by four. That's one way to think about it. So I could say, plus 1/8
times four cosine of four x. I'm just doing this so that it's... In theory you could do u-substitution, but we've have a lot
of practice with this. I have a function, I have its derivative, I could just integrate with respect to the four x right over there. I guess you could say it's
the reverse chain rule, which is really what
u-substitution is all about. Now I'm ready to integrate. So this is going to be equal to, I think we deserve a
little bit of a drumroll, 1/4 times 3/2 x, c-derivative of two x is
sitting right over here too, so this is going to be
minus sine of two x. We can verify that this
is a c-derivative of this. It's going to be two cosine of two x, we have it right over there, plus 1/8 times sine of four x. Derivative of sine of four x is going to be four cosine of four x, which is exactly what we have there. And then home stretch, we just write the plus C, plus sub constant. This is an indefinite integral. And we're all done. This wasn't the simplest of problems, but it also wasn't too bad. And it's strangely
satisfying to get it done.