Original subtitles

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Today we're going to be talking about how to use you substitution and then integration by parts to evaluate

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an integral.

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And in this particular case we've been given the definite integral of theta cube times cosign of theta

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squared d theta and we'll be evaluating that on the limits of integration.

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Square root of pi over to the square root of pi.

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I've gone ahead and written the integration by parts formula over here on the right because we're going

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to need that in a little bit.

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But this problem asks us to make a substitution first and then use integration by parts once we've simplified

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the functions inside of our integral.

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So we need to identify the substitution that we're going to make in this function and it should be at

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least a place to start with that you try to substitute for theta squared because having theta squared

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inside of our trigonometric cosigned function here means that we can have a little bit of trouble with

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this.

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We want to simplify what's inside our trigonometric function as much as we can.

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So let's make a substitution for theta squared and see where that gets us.

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In this case normally I would use you as the variable for substitution because it's use substitution

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most commonly.

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But we're going to need to have you available to us later for our integration by parts formula.

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So let's actually go ahead and use like X substitution and make a substitution for x in terms of theta

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instead of you so that we don't get confused later.

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So we'll call X theta squared and that will be the substitution that we're going to make.

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Remember that when you're dealing with you substitution you then take the derivative of what you just

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identified as X here and we'll call that dx so DX will be 2 theta.

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And then of course we add to this D theta.

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And now we want to solve for d theta we'll divide both sides by two theta and we'll get the theta is

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equal to dx over to theta.

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So now we can plug these values back into our integral so we'll end up with the integral with these

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limits of integration.

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And essentially what we have here is we have theta cubed and then we have cosign.

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Remember we substituted X for theta squared.

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So now we just have cosign of X and detailer we know is D X divided by two theta.

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The first thing we need to realize here is that we can cancel a theta from the numerator and denominator.

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So we'll go ahead and cancel out theta here.

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This will leave us with just theta squared in the numerator so we can go ahead and call this here theta

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squared instead.

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But remember that data squared.

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We set equal to x.

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So instead of theta squared here we will actually have is just x.

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And keep in mind that we have this this one half here the two in the denominator that we can move outside

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of the integral.

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So what we end up with before we start our integration by parts as we move the one half out in front

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of the integral.

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So we have one half our limits of integration and then what we're left with is just x cosign of x x

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and now we're in a great position to start integration by parts because we have two components inside

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of our integral.

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We have X and we have cosign of X so we need to identify you and DVH.

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Well in this case it's really obvious what you should pick for you because remember we want to pick

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something for you that will become simpler when we take its derivative d u and the derivative of X is

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just one which is much simpler than X so if we set you equal to x that means that divi has to be everything

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else inside of our integral.

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So divi automatically becomes everything else which is cosigned XTi X. So cosign x x.

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Now we take the derivative of you to get d u.

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So do you as equal to the derivative of X is just one.

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So we would have one DX which is just dx and then we take the integral of divi to get V.

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So we get a V equals the integral of cosign is sine.

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So we get sign of X and now we have all four of our components we can plug these into our integration

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by parts formula.

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So remember our integration by parts formula.

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Essentially we're going to be replacing this integral here this entire piece with what we plug in to

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the right hand side of our integration by parts formula over here.

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So we still need to include this one half that's out in front.

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So we'll get one half and then we'll multiply this by the right hand sidebar formula which is you Times

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V minus the integral of V times d u.

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And we have those components here.

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So we just plug in u and v US X and V is sign of x.

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So you Times V is just x sign of X and then we subtract the integral and again we're just following

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here our integration by parts formula V and D U.

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So we have V as sign of X and D u as DIAK.

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So we're just left with sign of x dx and remember that because we're dealing with a definite integral

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here we have these limits of integration these limits of integration apply to everything inside these

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parentheses here these big brackets.

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So we have to remember both this integral here and the sine x.

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So we have to remember to evaluate at the square of Pi over to 2 the square root of pi.

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Now thanks to integration by parts we have an integral that's really manageable we just have the integral

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of sign of X.

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Well we know that the integral of sign of x is negative cosign of x.

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So what we end up with here is 1 1/2 times x sign of X and because we're going to get minus and then

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negative cosign of x that will be plus cosign of X and that is our integral evaluated.

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And of course we have our limits of integration here that we'll have to evaluate this function at.

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But before we do that now that we've finished taking the integral of everything we have no integrals

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left we need to go ahead and back substitute for X.

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Remember we said that X was equal to theta squared.

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Well now that all our integrals are gone it's time to put X back in terms of theta.

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So what we'll have is one halftimes theta squared times sign of theta squared plus cosign of theta squared.

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And now we want to evaluate on these limits of integration and where we plugging these limits of integration

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in for theta.

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Remember when you're evaluating at limits of integration like this you always plug in this top number

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first and then subtract whatever you get when you plug in the bottom number.

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So we'll plug in the top number which is square of Pi.

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First and what we'll get is the square root of pi squared.

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Well a square root squared takes away that square root.

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So the square root of pi squared is just pi.

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So we end up with pi times sign of data squared.

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And again the squared of Pi squared is just pi.

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So we end up with sign of pi plus cosign obviously of Pi again.

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And then we'll subtract whatever we get when we plug in the lower limit of integration.

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So we need a parentheses here so that this negative sign applies to everything we've got in here.

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So we'll get the square root of pi over two squared which will just give us of course PI over 2 and

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then we'll have sign of Pi over to plus cosign of Pi over to.

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So now we just evaluate We have 1 1/2 times sign of Pi right here.

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Sign of Pi is zero.

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So zero times pi will obviously just give us still 0.

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This is going to go away completely cosign of Pi is negative 1.

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So we've got a negative one there than we have here minus sign of Pi over 2 is 1 so 1 times pi over

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2 is Pi over 2.

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And then cosign of Pi over two is zero.

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So we just have a minus zero.

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We don't need to account for that.

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So we just have negative 1 minus PI over 2.

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Now if we distribute the 1 half we'll get negative 1 half minus PI over 4 and that's it.

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That's our final answer.