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

$ \displaystyle \int_{-\pi}^{\pi} \sin mx \sin nx dx = \left\{

\begin{array}{ll}

0 & \mbox{if $ m \neq n $}\\

\pi & \mbox{if $ m = n $}

\end{array} \right.$

$m+n, m-n$ are integers

Integration Techniques

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University of Michigan - Ann Arbor

Idaho State University

Boston College

here we like to prove the following formula for imager valued and an end, the integral from negative fight, a pi sign MX Time sign and X is either zero or pie, depending on whether or not emon enter equal. So let's consider two cases here. Let's consider the first case in which M equals N and we can write. The cinder girl is negative. Pika pi Sign a mix times sign a mix again. This is because for taking em to be in so we have the integral negative Pika pi sine squared MX And at this point, it will help us to use the following identity we can always right Science square of t is one minus coastline of two tea all over too. So let's use this formula here where we have t equals MX in our case. So using this pulling out the one half negative pira pie one minus co sign to a mix and we can integrate this if this too immense is giving you a hard time. You can go out and using U substitution. So it's evaluate this integral. We have one half X minus sign to MX over to him and her and points are negative. Pika pi suspending plug was in So we have pie minus sign soup. I am over to win. This is that this is after we plug in pie and now we plug in negative by So we have minus pi for X minus Sign negative two pi am Now we were given that m and enter in Is your value so well here and two pi m za energy or multiple of two pi And we no sign of any multiple of two pi any image or multiple super zero So these two terms go to zero and we're left with one half ofthe pie minus negative pie just too high. Did you cancel off those twos to get pie? And that verifies the formula in the case here when Eman enter evil. So let's go to another page to consider the case when mnr distinct So this is case too and since our inner grand is of this form so it's right there in a girl here we have negative pi pi sign MX Sign Annex Let's use a formula for the product of signs giving it given in the textbook in this section sign A sign be is co sign a minus B minus co sign of a plus B all over, too. Sonar problem. Let's identify the A and the B is MX bia's annex. So, applying this formula, we have one half integral. Now we have co sign of a minus me. So let it be co sign of M X minus annex minus co sign of a plus B. Next thing we should do is just simplify these factor on X from the inside We have one half in a girl Negative pika pi co sign and then we have ah, and minus and x minus co sign M plus and parentheses. X the ex and we could actually go ahead and evaluate these intervals. If the's m minus and templates and her throwing you off, feel free to use the U substitution here You could take you to be a minus in its and for this one you could take you to be and plus and X So it's evaluate these We have one half sign of and minus in its over on minus end, minus sign of M plus annex over implicit. Oops. And here we have our own points negative pileup. I also here there's an observation to be made that why we weren't able to use this approach last time. Here we don't have to worry about the denominators being zero This one's non zero because M is not equal to end and here and plus and is not zero since m unendurable positive that was given assumption. And this is the reason, especially this first one over here, the M minus and in case one lemon and we're equal. So we would've had a zero on the bottom. That's why we had to use a different method. So now let's go ahead and evaluators. So we'LL plug in Hi friends one half sign and minus and pie over and minus and minus Sign of M plus and pie over unpleasant. So this is after plugging in pie. Now we plug your negative pie and now we observe that since Emin and managers we also have that EMP listen and minus and our insiders No. So that means that in all of these signs these are all insecure multiples of pie. And we know from the unit circle that if you take a manager most for pie. The sign is zero there. So all four of these expressions or zero So we have one half time zero, which was zero. And so this also proves the formula in the case in which a man enter, not equal. We've resolved both cases, so this completes approved.