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JH
Numerade Educator

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Problem 69 Hard Difficulty

Prove the formula, where m and n are positive integers.

$ \displaystyle \int_{-\pi}^{\pi} \cos mx \cos nx dx = \left\{
\begin{array}{ll}
0 & \mbox{if $ m \neq n $}\\
\pi & \mbox{if $ m = n $}
\end{array} \right.$

Answer

$\frac{1}{2}$

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Video Transcript

here we'd like to prove the formula that the integral from negative pika pi of co sign of MX Times Co sign of N IX is either zero reply depending on whether or not MNR equal. So here we should break this. Since the two cases, it's considered the first case in which imminent or the same and let's write the integral we have co sign MX and now since MNR equal, we have another co sign MX so we can go ahead and write this So we have co sign squared and next the eggs. And now at this point, we should use an identity to simplify. We can use the fact that co sign squared t is one plus coastline to tea all over too. So if we apply that here with T equals MX, we'LL have one half in a real negative Pika pi one plus co sign of two women IX and we're ready to but in a great peace to know if you're having a hard time with this to him in front of the eggs, feel free to using substitution to make it a little easier. So we have one half and a girl one is X integral of co signed to our necks is sign Swim X all over to him and then our end points are negative bite of pie so it's bottom club goes in. What pie plus sign of soup I am So this is after plugging in pie and we plug it negative pie well minus pi plus sign negative soup I am now. Here's where we used the assumption that M is a manager. So two pi ems, that energy or multiple of pie. So using our knowledge of the struggle, we know that these signs from zero and we're left over with one have a pie plus pie, which becomes one half of two pi, which is just pie. And this verifies the formula in the case in which eminent are equal. So let's separately consider the case in which and then enter distinct. So it's for a new page. Now we're in case, too, and then you're not evil. So in this case, let's write in a roll Cose Dynamics Coastline Annex. Now here it will be helpful to use a formula for the product of two co signs given in this section close I name times close. I'm bi. His co signed a minus plus co sign of a plus B, all of it. So let's use this formula where here in our problem is immense. He's an ex applying this identity. We have one half in a roll, negative kind of high. Then we have co sign of a minus. So an X minus Alex plus co sign of a plus. And that's plus annex. Now, to make this easier to integrate, you should fasten our exes from the fantasies. Okay, now we're ready to integrate. It might help you. If the's and minus and eggs in the M plus an extra throwing you off, you can go ahead and use up here for this one. We could take and plus index so penetrating these we'LL have sign of and minus and over and minus and plus sign of employees in X all over implicit, you know, and points are negative. So let's make some observations here for about regarding the denominators so and minus and is not zero says em is not equal to end. This is by our assumption and in this case, and for the other denominator and plus in is not equal to zero, since they've given assumption in the problem. Was that Damon and more? Both positive. So thes denominators or both Down zero, That should always be changed. So now we're ready to plug in the end. Points exploding pie. First, we have sign of a minus and pie a minus in, plus sign of them. Plus in pie over implicit. So this is from plugging in pie. And now we plug in negative fire. Hi there. On minus and a sign of and plus in negative five all over in person. All right, so here, let's make some more observations and write this up here What? And plus in and minus and our interviews. So that means here and minus and in an plus and pie, these air, all imager, multiples of pie And we know from sign with are in the circle that these are all zero. So zero zero zero zero. So the final answer is one half times zero, which is zero. And this proves the formula and the other case in which em in and are not equal. And that completes the proof