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## Integral Calculus (2017 edition)

### Course: Integral Calculus (2017 edition)>Unit 6

Lesson 5: Integration using trigonometric identities

# Integral of sin^4(x)

It is a bit involved, but we can use u-substitution to find the integral of sin^4(x).

## Want to join the conversation?

• Why does sin^2(x) = (1/2)(1-cos(2x) not (1-cos^2(x)) ?
• It does, however converting from one trig function that is squared to another that is squared doesn't get you any further in solving the problem. But converting a squared trig function to one that isn't squared, such as in the video, well, sin²x gets you 1/2 - cos(2x)/2, and that you can integrate directly.
• At around , why does (1-cos(2x))² turn into 1-2cos(2x)+cos²(2x) ? Where does the 2cos(2x) come from ?
• Sal was simply expanding the expression (1/2 (1-cos2x))^2. 1/2 squared is 1/4 and (1-cos2x) squared is (1-cos2x) times (1-cos2x). If you recall from Algebra that (a-b)(a-b) = a^2 -2ab + b^2, then let a = 1 and let b = cos2x and multiply it out. a^2 = (1)(1) = 1, -2ab = -2(1)(cos2x) = -2cos2x. b^2 = (cos2x)(cos2x) = cos^2 2x. Hope this helps. Good luck.
• Do i have to memorize the double angle identity formula? I searched the trig identities formula. But there are so many.. :(
• (In case anyone wanted) A quick derivation for the trig identity {sin^2x=1/2(1-cos2x)} :
2. Rearrange both: sin^2x=1-cos^2x and cos^2x=cos2x+sin^2x
3. Substitute cos2x+sin^2x into sin^2x=1-cos^2x for cos^2x
4. Expand: sin^2x=1-cos2x-sin^2x
5. Add sin^2x to both sides, giving 2sin^2x=1-cos2x
6. Divide both sides by 2, leaving sin^2x= 1/2(1-cos2x)
• At , how do you get (1/8) *4cos4x ??
• Hi @redthumb.liberty, is ∫( 1/2*cos(4x) )dx = ∫( 4/8*cos(4x) )dx reverse chain rule?
(1 vote)
• Where does the Trig identity at come from?
• Please, can anyone explain how (sinAx)(sinBx) = 1/2(cos(Ax-Bx)-cos(Ax+Bx)) Thanks
• Difference Formula:
cos(Ax - Bx) = cos(Ax)cos(Bx) + sin(Ax)sin(Bx) *
Sum Formula:
cos(Ax + Bx) = cos(Ax)cos(Bx) - sin(Ax)sin(Bx)
<=> - cos(Ax + Bx) = - cos(Ax)cos(Bx) + sin(Ax)sin(Bx) **
Add equations * and **. We get
cos(Ax - Bx) - cos(Ax + Bx) = 2sin(Ax)sin(Bx)
<=> sin(Ax)sin(Bx) = ½(cos(Ax - Bx) - cos(Ax + Bx))
• Is
sin^2(x)=(1-cos(2x))/2
&
cos^2(x)=(1+cos(4x)/2
another two trig function properties? I've never met them before and wonder if anyone can provide me the link to these properties, either articles or vids.
• In my trigonometry book, there is the following identity:
sin P + sin Q
= 2sin((P + Q)/2) cos((P – Q)/2)
and also one for sin P – sin Q and cos P ± cos Q. It's mentioned that this identity can be helpful in integration, but I can't see how this identity can be helpful, especially in integration. Can someone please tell me a case where this could be helpful? Thanks in advance.
• Those are sum to product identities. I agree with you, at first glance it is hard to imagine a situation that is simplified by changing a sum to a product - but there could be - it is always best to have these things in your toolkit, just in case.
It is usually most useful going the other way, turning a product, such as cos(Px)sin(Qx) into a sum, eg:
∫sin(3x)cos(2x)dx = ½∫(sin(5x) + sin(x))dx = ½∫sin(5x)dx + ½∫sin(x)dx
• at i dont understand why the 2 disappeared after you integrate the 2 cos2x which ibecamesin 2x..same as 4cos4x which became sin4x.
• my teacher is asking for the integral of (secx)^4; how do we solve for secant and cosecant to an even power?

## Video transcript

- [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.