Question Video: Finding the Integration of the Product of Two Sine Functions Having Different Angles Using Trigonometric Identities Mathematics

Video Transcript

Determine the integral of negative eight sin four 𝑥 sin nine 𝑥.

So then, the first thing we can do is we can take out our constant term. And that’s negative eight. So, now we got negative eight multiplied by the integral of sin four 𝑥 sin nine 𝑥. And then what we can notice is we’ve got one of our trig identities. So, we can apply the product-to-sum formula.

And that is that if we’ve got sin 𝑥 multiplied by sin 𝑦, this is gonna be equal to a half cos 𝑦 minus 𝑥 minus cos 𝑦 plus 𝑥. So therefore, if you apply this, we’re gonna get negative eight multiplied by the integral of a half cos nine 𝑥 minus four 𝑥 minus cos nine 𝑥 plus four 𝑥. So therefore, we’re gonna get negative eight multiplied by the integral of a half cos five 𝑥 minus cos 13𝑥.

So then, again, what we can do is take the constant term out. So, we’ve now got negative four multiplied by the integral of cos five 𝑥 minus cos 13𝑥. And to integrate this, what we’ll do is we’ll split it up. So, we have the integral of cos five 𝑥 minus the integral of cos 13𝑥. So, to integrate cos five 𝑥, first we’ll substitute 𝑢 for five 𝑥. And then we’ll differentiate. So, we’ll differentiate to get d𝑢 d𝑥.

So, if we differentiate five 𝑥, we’ll just get five. So therefore, we’ve got d𝑢 d𝑥 is equal to five. And the reason we’re doing this is so that we can rearrange because we want d𝑥 in terms of d𝑢. And when we do that, we get d𝑥 is equal to one over five d𝑢. So, now we can put this back into our integral. So, we now have a fifth multiplied by the integral of cos 𝑢 d𝑢.

Well, this is much more straightforward because cos 𝑢 is just gonna be one of our standard integrals. And that’s because the integral of cos 𝑥 is just sin 𝑥. So therefore, the [integral] of cos five 𝑥 is gonna be equal to sin 𝑢 over five, which is equal to sin five 𝑥 over five cause we have substituted back in the 𝑢. So, that’s the first part of our integral complete. So, let’s move on to the second part because we’ve got the integral of cos 13𝑥.

Again, we’re gonna use the substitution method. And we’re gonna sub 𝑢 is equal to 13𝑥. So, d𝑢 d𝑥 is gonna be equal to 13 cause we differentiate 13𝑥. So therefore, in terms of d𝑢, d𝑥 is equal to one over 13 d𝑢. So, now what we need to do is integrate cos 𝑢 as we did before. And when we integrate cos 𝑢, we’ll get sin 𝑢. So, we’ve got sin 𝑢 over 13. And then we substitute back in our 13𝑥 for 𝑢. So, we get sin 13𝑥 over 13. So, that’s the integral of cos 13𝑥.

So then, we put this back into our integral. So, we’ve got minus sin 13𝑥 over 13. And then, we’ve got plus 𝑐 on the end cause we can’t forget our constant of integration. So, now all we need to do is multiply through by negative four. So therefore, we get negative four over five sin five 𝑥 plus four over 13 sin 13𝑥 plus 𝑐. And we got plus four over 13 sin 13𝑥 because we have negative four multiplied by negative sin 13𝑥 over 13. Negative multiplied by negative is a positive.

So therefore, we can say that we’ve determined the integral of negative eight sin four 𝑥 sin nine 𝑥. And we’ve done that using one of our trig identities and the product-to-sum formula. So, we’ve got the result as negative four-fifths sin five 𝑥 plus four over 13 sin 13𝑥 plus 𝑐.