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Integration of powers and products of sine and cosine.
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Powers of sine and cosine.
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The general problem we'd like to address here is integration of the
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Mth power of sine x times the Mth power of cosine x dx,
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where M and N are non-negative integers.
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Of course, the simplest cases here are where M equals 0 and N equals 1,
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and we have the indefinite integral of cosine x dx equals sine x plus c,
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and the case where N equals 0 and N equals 1,
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and we have the integral of sine x dx equal to minus cosine x plus c.
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Now let's look at an example of a little bit more complicated situation.
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Suppose we want to compute the indefinite integral of cosine cubed x dx.
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The key here is to notice that we can take one of the cosine factors
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and group it with dx, and think of that as the du that arises from
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a substitution of u equals sine x.
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Now in order to make that substitution, we need to express the rest of the
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integrand in terms of sine.
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That's easy to do for cosine squared x, because that's equal to 1 minus sine squared x.
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So now we make the substitution, and we have 1 minus u squared du,
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where the integrand is just a simple polynomial.
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The integration now is easy. We get u plus 1 third u cubed plus c,
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and before we put u equals sine x back in, let's factor this and write it as 1 third of u times 3 minus u squared plus c.
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So we end up with 1 third of sine x times 3 minus sine squared x plus c.
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Now let's think about what it was about this integrand that allowed this method to work.
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We factored off the first power of cosine x to group with dx,
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the idea being that we were going to substitute u equals sine x.
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Then we needed to express the rest of the integral in terms of sine.
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We were able to do that because we had an even power of cosine there,
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which made the replacement of cosine squared x with 1 minus sine squared x
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result in a very simple integrand.
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So the key to all of this was the fact that after we factored out cosine x,
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we were left with an even power of cosine.
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Now this happened precisely because the power of cosine that we started with was odd.
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And we will see that whenever we have an odd power of sine or cosine,
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the same general approach will work.
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Now let's look at a similar example.
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Let's calculate the indefinite integral of cosine squared x times sine cubed x dx.
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Here, since we have an odd power of sine, we can factor off the sine to the first power
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and group that with dx.
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Now notice that sine x dx is actually the negative of du if u is equal to cosine x.
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So we'd like to make a substitution of u equals cosine x here.
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Now in order to do that, we have to express everything in terms of cosine.
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So we'll replace sine squared x with 1 minus cosine squared x.
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Now we make our substitution, u equals cosine x du equals minus sine x dx.
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And we get the integral of u squared times 1 minus u squared times minus du.
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Now to do the integration, we'll expand that product and write this as the integral of u to the fourth minus u squared times du.
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Now we integrate and get 1 fifth u to the fifth minus 1 third u cubed plus c.
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And before we do the replacement here, let's factor. We can factor out a 1 fifth u cubed.
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That makes the other factor 3u squared minus 5.
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And now we replace u with cosine x.
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So in general, we can integrate products of powers of sine and cosine in more or less the same way as long as we have at least one odd power.
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If the power of cosine is odd, that is, the exponent is of the form 2k plus 1.
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We factor off cosine x dx and think of that as du for the substitution u equals sine x.
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That means we have to express the rest of the integrand in terms of sine.
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Cosine to the 2k power is just cosine squared raised to the kth power.
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So we replace cosine squared with 1 minus sine squared.
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Now making the substitution u equals sine x du equals cosine x gives us the integral of u to the m times 1 minus u squared all to the kth power du, which is an principle a simple thing to integrate because it's just a polynomial.
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On the other hand, if power of sine is odd, then we group a factor of sine x with dx with the idea of substituting u equals cosine x.
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So we need to express the even power of sine in terms of cosine.
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So we replace sine squared with 1 minus cosine squared, and then make our substitution of u equals cosine x and du equals minus sine x dx.
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And again, we get the integral of a polynomial in u.
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Now let's look at an example that indicates that the same approach can be used in slightly more general problems in which the powers aren't necessarily positive.
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So consider the integral of cosine cubed x divided by sine x dx.
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Note that this integrand might also be expressed as cotangent x times cosine squared x, or even cosine cubed x times cosecene x.
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But since we have an on-power of cosine in the numerator, we can group a factor of cosine x with dx, and then attempt to make the substitution u equals sine x.
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In order to do that, we simply need to express the rest of the integrand in terms of sine.
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So we replace cosine squared x with 1 minus sine squared x.
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And now we make our substitution, and we get 1 minus u squared divided by u du. To do the integration here, we need to split this fraction up into two terms.
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The integral of 1 over u minus u du, the 1 over u term gives us natural log of the absolute value of u, and minus u gives us 1 half of u squared plus c.
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Now u is equal to sine x, so we end up with the natural log of the absolute value of sine x minus 1 half sine squared x plus c.
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So now we need to address the situation where the previous method doesn't work.
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That's the situation where sine and cosine are both raised to even powers. One way of handling even powers is to use power reduction formulas.
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Let's recall from trigonometry, the identities cosine squared theta equals 1 half of 1 plus cosine 2 theta, and sine squared theta equals 1 half of 1 minus cosine 2 theta.
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Now let's look at the two simplest cases. First, let's consider the integral of cosine squared x dx. By the identity above, this is just 1 half of the integral of 1 plus cosine 2x dx.
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This gives us 1 half of x plus sine of 2x divided by 2 plus c.
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In some situations, we would prefer to write the result in terms of cosine x and sine x rather than sine of 2x. So we can use a double angle formula here that says sine of 2 theta equals 2 sine theta cosine theta to rewrite our result as 1 half of x plus sine x times cosine x plus c.
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Now let's look at the integral of sine squared x dx. Notice that the identity for sine squared theta is just like the identity for cosine squared theta except there's a minus sign in the middle.
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So we basically end up with the same result except with a minus sign in the middle. We have 1 half of x minus 1 half of sine 2x plus c.
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Or applying the double angle formula, we have 1 half of x minus sine x times cosine x plus c.
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Now let's look at a slightly more complicated example involving even powers. Let's integrate cosine squared x times sine squared x dx.
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Let's do this by using our power reduction formulas. We replace cosine squared x with 1 half of 1 plus cosine 2x and replace sine squared x with 1 half of 1 minus cosine 2x.
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Now let's multiply out those factors. We'll pull 1 fourth outside and the product of 1 plus cosine 2x times 1 minus cosine 2x is 1 minus cosine squared 2x.
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Now we'll handle cosine squared 2x in a similar way. We'll apply the same power reduction formula and write that as 1 half of 1 plus cosine of 2 times 2x or cosine 4x.
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Now let's simplify the integrand. Notice we have 1 minus 1 half for 1 half and then minus 1 half of cosine 4x. So we can factor out a half and write this whole thing as 1 eighth times the integral of 1 minus cosine of 4x dx.
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Now the integration is easy. We get 1 eighth of x minus sine of 4x divided by 4 plus c. Now let's factor out 1 fourth and write this as 1 over 32 times 4x minus sine of 4x plus c.
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Now if we want to write sine of 4x in terms of cosine x and sine x, we need to come up with a quadruple angle formula. That's actually quite easy to do.
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We'll first apply the double angle formula here to sign of 2 times 2x, writing it as 2 times the sine of 2x times cosine of 2x.
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Next we apply the double angle formula for sine to write 4 times sine x times cosine x and then rewrite cosine of 2x as cosine squared x minus sine squared x.
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Now notice that if we replace the sine of 4x with this and our result, both terms in the parentheses have a factor of 4 which we can bring out front and we have 4 over 32 or 1 eighth times x minus sine x cosine x times cosine squared x minus sine squared x plus c.
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In general, when we have even powers of sine and cosine, more or less the same process can always be carried out. We apply power reduction formulas, then expand products and then repeat that process until the integrand involves only on powers of sine and cosine.
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Even powers by parts, let's look at an example of how we do a fairly simple case by parts. Let's consider the integral of sine squared x dx.
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How do we approach this by parts? Well, we'll let u equal sine x and let dv equal sine x dx. That means that du is cosine x dx and v is minus cosine x.
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And now uv is minus sine x times cosine x and v times du is minus cosine squared x dx. So we have minus the integral of minus cosine squared x dx.
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Now let's rewrite cosine squared x as 1 minus sine squared x. Now we can break that integral into 2 pieces. We have the integral of dx which gives us some x and then minus the integral of sine squared x dx.
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So now notice that we have the integral we began with on the right hand side with a minus in front. So let's add the integral of sine squared x dx to both sides.
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Then we have 2 times the integral of sine squared x dx on the left and minus sine x times cosine x plus x plus c on the right.
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Now we simply divide by 2 and we have our result. One half of x minus sine x times cosine x plus c.
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In a similar but slightly more complicated way, we can prove the following reduction formulas.
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The integral of the nth power of cosine x dx equals 1 over n times the n plus 1st power of cosine x times sine x plus n minus 1 over n times the integral of bn minus 2 power of cosine x dx.
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And we also have a very similar formula for the integral of the nth power of sine.
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Now let's look at an example where we use those formulas. This integrate the 4th power of cosine x times sine squared x dx.
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Here we'll make use of the reduction formula for powers of cosine.
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The first thing we'll do is replace sine squared x with 1 minus cosine squared x and then multiply out that product and break the integral into 2,
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the integral of cosine to the 4th x dx minus integral of cosine to the 6th x dx.
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Now let's apply the reduction formula to the integral with the highest power.
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We'll leave the first integral alone for now and applying the formula to the second term we have minus 1 sixth cosine to the 5th x times sine x plus 5 over 6 times the integral of cosine to the 4th x dx.
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Now let's combine those two integrals. Each one is an integral of the 4th power of cosine so we have 1 minus 5 sixth or 1 sixth of the integral of cosine to the 4th x dx.
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And now we apply our formula to that integral with n equals 4.
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We have 1 fourth of cosine cubed x times sine x plus 3 over 4 times the integral of cosine squared x dx.
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Now we can apply the reduction formula to the integral of cosine squared or it's a fairly simple matter to approach it using the power reduction formula.
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But in either case we end up with 1 half of cosine x times sine x plus 1 half times x for that integral.
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So now let's clean this up a bit.
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Notice that what's in the inner parentheses gives us a factor of 1 half there for a factor of 3 eighths outside those parentheses.
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So let's go ahead and factor a 1 eighth from the whole thing. Then we have 1 sixth times 1 eighth or 1 over 48 times the quantity in parentheses here.
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Now notice that every term in parentheses except 3x has the factor cosine x times sine x.
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So let's go ahead and do that partial factorization.
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Products of factors with different periods.
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Here we're going to consider problems of the following forms.
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Integrans involving a product of 2 signs or 2 cosines or a sine and a cosine where the 2 factors have different periods.
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That is there's a different coefficient of x inside of each of the 2 factors.
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The key to dealing with these is to use trigonometric product to some formulas.
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Let's look at an example.
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Suppose you want to integrate sine of 5 pi x times sine of 2 pi x dx.
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The integrand here has a graph that looks like this.
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Now since the integrand has to form sine a times sine b, we'll apply the first of these products to some formulas to rewrite our integrand as 1 half of cosine of 5 pi x minus 2 pi x, which is 3 pi x,
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minus the cosine of 5 pi x plus 2 pi x, which is 7 pi x.
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Now we can antideifferentiate each term separately.
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The first term will give us a sine of 3 pi x divided by 3 pi.
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And the second term gives us minus sine of 7 pi x divided by 7 pi.
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Now let's factor out the least common denominator there, 21 pi.
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And we have 1 over 42 pi times 7 times sine 3 pi x minus 3 times sine of 7 pi x plus c.
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Now let's summarize everything we've seen here.
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For the problem of integrating the nth power of sine x times the nth power of cosine x dx, we've seen that if we have an odd power of cosine,
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we can substitute u equals sine x and du equals cosine x after grouping cosine x with dx and then expressing the remainder of the integral in terms of sine x.
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If we have an odd power of sine, then we make the substitution u equals cosine x and du equals minus sine x dx after grouping sine x with dx and expressing the remainder of the integral in terms of cosine x.
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If both powers are even, then we have two ways of approaching things. We can either reduce powers with half angle formulas or use integration by parts.
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Reduction formulas encapsulating a step of integration by parts are very useful there.
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Also, for processing our final result, we want to remember double angle formulas sine of 2 theta equals 2 sine theta cosine theta and cosine 2 theta equals cosine squared theta minus sine squared theta.
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And we've also seen that when we have products of factors with different periods, we can apply product to some formulas to convert our integrand into something that's easily integrated.