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Numerade Educator



Problem 3 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int_0^{\frac{\pi}{2}} \sin^7 \theta \cos^5 \theta d \theta $




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Video Transcript

This problem is from Chapter seven section to problem number three of the book capitalists Early Transcendental Sze, eighth Edition by James Store Here we have a definite integral from zero a pie or two Science of the seven power of data Coastline to the fifth power of data. So since we have ah, odd power on the co sign, we can rewrite this by pulling on one founder. Of course. So we re read co signs the fit as co sent to the fourth Power Times Co sign. Okay, so observe here We have a co sign to the fourth power so we can rewrite this coastline to the fourth is simply co sign squared, squared. Now, applying a pit Agron identity I can rewrite. This is one minus sine squared, squared and then evaluating the square at the end, I have one minus to sign square, plus signs of the four power of theater. So are integral becomes the integral zero for two signs of the seven power. And then now we replaced coastline to the fourth power with our latest expression over here. So we have ah one minus two cents for data plus signs of the fourth power of data. And let's not forget our co signed Ada di Beta at the end. So here we see that we should apply u substitution here, let's take you two be sign of data then do you disco Santa d data. And that's exactly what we have in our interval. Also here because we're dealing with a definite integral we're we'll have to change the bounds of integration. So originally we were starting the animal a zero. So that means you will started zero as well because sign of zero zero And since data was goingto pie over too you goes toe put a sign up high over too. Bush is one. So these are new bounds of integration. After we make the switch from data to you, you goes from zero to one and now we read, Write everything in terms of you on Lee. So we have ah you to the seventh one minus to you square. Plus you know the fourth deal. So before we integrate, let's distribute tissue to the seven inside the parentheses. So where were you? To the seventh Power minus to you to the ninth, plus you to the eleventh hour to you. So here we'LL proceed by applying the power all three times. So for the first time, we have Ah you today over eight minus to you to the ten all over ten plus you to the twelve hour over twelve. No need for a constant integration because we have a definite rule from zero one. So now we plug in one and zero and for you and subtract. So first we plug in one. So once a day it is just one minus two times one over ten. So to Overton, plus one over twelve. So this is from plugging in one, and now we're plugging zero. But when we plug in zero for you, each of these three terms goes to zero, so we just subtract zero. And lastly, we could proceed by just finding a common denominator here. And so one way to do this is to just put everything over one hundred and twenty. Someone would do that. So here will multiply top and bottom. My fifteen top and bottom of this by twelve. So two times twelve and top and bottom of this by ten. So simple. Find this. We got our answer, which is one over one hundred twenty