Exercises 59–64 require the use of various trigonometric ident...

Question

Exercises 59–64 require the use of various trigonometric identities before you evaluate the integrals.

∫ sin(θ) sin(2θ) sin(3θ) dθ

Accepted Answer

Welcome back everyone. In this problem we want to evaluate the integral of sin 2 x multiplied by sin 4x multiplied by sin 6 X with respect to x. Now to help us first simplify our product here to make it easier to integrate, we can apply the product to some identities. Now recall here that for the product of two sine functions, sine A and sine B. The product will be equal to 12 of the cosine of a minus b minus the cosine of a plus b. So we can apply this to first rewrite the sin of 2 x multiplied by the sin of 4 x, and then we can multiply our result by the sin of 6 x to help us simplify. So basically we can apply our product to some identity twice. So let's first start by simplifying the sin of 2 x multiplied by the sin of 4 x. So using the identity. Then the sine of 2x multiplied by the sine of 4 x. Is going to be equal to half of the cosine of 2 x minus 4 x. Minus the cosine of 2 x + 4 x. I know when we simplify that, it will be a half of the cosine of -2 X, OK, minus the cosine of. 6 X, OK. Which will be equal to 12 of the cosine of 2 x. Minus the cosine of 6 x, because remember, the cosine is an even function, so the cosine of -2 x is the same as the cosine of 2 X. Now we need to multiply this result by the sin of 6 x again, by using our using our product to some identity, OK, so here, let me put this in red. Thus, first we're rewriting the sine of 2 x multiplied by the sine of 4 x. Multiplied by the sign of 6 x, OK. And we're saying that this is equal to, based on what we just found earlier. I have. OK. Off. The cosine, well, let me write this out properly here. A half of the cosine of 2 x minus the cosine of 6 X, OK. Multiplied by the sine of 6 x. So to apply our products of identity, first we'll need to multiply the cosine of 2 x by a sin of 6 x and the cosine of 6 x by the sine of 6 x. Now notice it's a little different here because no we are multiplying cosine terms by sine terms. So again, recall. That to do that, the, in other words, if we're multiplying the sin of a by the cosine of b, it will be equal to a half of the sin of a plus b. Plus the sin of a minus b. So now if we apply that idea. OK, then for our first term, the cosine of 2 x multiplied by the sine of 6 x. It's going to be equal to half of the sin of. 2 X plus 6 X, which is 8 X, OK, plus the sign of 2 x 6 X minus 2X, sorry, which is going to be 4 x, OK. So that would be the cosine of 2 x multiplied by the sine of 6 x. On the other hand, the cosine of 6 x. OK, multiplied by the sin of 6 x. is going to be equal to a half of the sin of 6 x minus 6 x, which is the sin of zero, OK, plus the sin of 6 x plus 6 x, which is 12 x, and now this would simplify to a half of the sine of 12 x. So now if we combine all of this together, OK, then our integral, let's put that in red here, our integral of the sin of 2 x multiplied by the sin of 4 x. Multiplied by the sin of 6 x with respect to x is now going to become the integral of a quarter of the sin of 4 x, OK. Plus a quarter of the sine of 8 X. OK. Minus a quarter of the sin of 12 x, all with respect to x, and now remember we got these quarters because earlier we were supposed to multiply everything by a half, OK, from the first time we applied our product to some identity. No, we need to integrate each term. Now if we think about it here, first for the inte we can spread our integral out means we're going to integrate each term separately and for starters. the integral of the sin of 4 x will be a quart negative quarter of 4 x. So in other words, for our first integral, it's going to be a quarter multiplied by a negative quarter of the cosine of 4 x, OK. For our second term, a quarter of the sin of 8x, that's going to be a quarter of -1/8 of the cosine of 8x, OK. And for our 3rd term. Yeah. g quarter multiplied by the sin of 12 x, that's going to be. Negative, sorry, a quarter, OK. Multiplied by -25. Of or sorry, -112. Rather, or rather let me write that properly here where we were subtracting my apologies, so that's minus 25 multiplied by negative. -112. Of the cosine of 12 X, OK. So these are our three terms and remember at the end we each of them would integrate to have a constant C1, C2, C3, so we just end up with a final constant C. Now if we were to simplify this, our first term would become 1/16 of the cosine of 4 x. The second term 1 divided by 32, multiplied by the cosine of 8x, and 4 or third term that becomes plus 1/48 of the cosine of 12 x. Thus this is going to be the integral of the sin of 2 x multiplied by the sine of 4 x multiplied by the sine of 6 x with respect to x. Thanks a lot for watching everyone. I hope this video helped.

Exercises 59–64 require the use of various trigonometric identities before you evaluate the integrals.∫ sin(θ) sin(2θ) sin(3θ) dθ

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Exercises 59–64 require the use of various trigonometric identities before you evaluate the integrals.
∫ sin(θ) sin(2θ) sin(3θ) dθ