Problem 4 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int_0^{\frac{\pi}{2}} \sin^5 x dx $

Answer

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Video Transcript

this problem is from chapters Heaven Section to Problem number four of the book Calculus Early Transcendental Sze eighth Edition by James Door Here we have a definite a rule from zero up IRA to science to the fifth hour of ex DX. So the first thing we can do here is just rewrite this in a roll by pulling out one factor of sign. So by doing this will have science of the fourth power Becks Times, Santa picks D x. So here we have a science of the fourth power. So let's go to the side and make some observations here so we can rewrite signed to the four sides, squared x in the parentheses and we can evaluate this insider thing. See, in one minus to co sign Square Plus co sent to the four power so are integral becomes the integral from Sierra over, too. And we just replace that into the fourth with our latest expression over here on the right, So have a one minus to co sign square plus coastline to the fourth power times sine x dx. And here we should apply u substitution. Let's think you two be cause and effect so that to you is negative Sign X the ex We don't have this negative in front of the sign d x in our original integral. So we'LL make up for it by just putting a negative and argue so negative to you Eagles sine x dx. Another observation here is that we have a definite no girl. So since we're making a u substitution, the limits of integration will change. So originally X goes from zero to Piper too. So now that will change in our new variable You. So the starting value for you will be cosan zero, which is one and the upper limit for you will be co sign the power to which is zero. So now let's rewrite are integral using our new variable you So we change those limits of integration from two one and zero and we have a one minus to you square. Plus you know, the fourth to you. And let's not forget that minus sign in the front because we wanted negative, do you? Not just to you. So here before we integrate, we could just push this negative inside. So have a minus one plus two. You square minus you forth, do you? And now we could apply the power or three times. So we have a negative. You plus two, you cued over three, minus your of the fifth power over five. And our end points are one and zero. So when we played in zero first and for you, each of these three terms becomes a zero. So we have a zero minus, and now we plug in one for you. So have a minus. One plus two over three minus one over five, which becomes one minus two or three plus one over five. And here we could get a common denominator of fifteen, twelve, fifteen minus. So here we have the multiplied happened bottom of two over three by five. So we're gonna tent up there and then here on the one over five would multiply top and bottom by three. All over fifteen are denominator. And then we simple by this to get in over fifteen, and that's our answer.