Original subtitles

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Today we're going to talk about how to use integration by parts to find the antiderivative of function.

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To complete this problem well understand the integration by parts formula match that this is our function

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to the parts of our formula and then evaluate the integral.

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In this particular video we've been asked to evaluate the integral of each of the 7 x times cosign of

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2 x x.

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Now since this is an integration by parts problem the first thing that we need to do is identify in

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our function a value for you and a value for.

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If we look at the integration by parts formula here what we can see is that it tells us that if we're

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taking the integral of you Times DVH then we can set that equal to you at times V minus the integral

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of the times d u.

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So what we have to first identify is value for you and a value for divi because the integral over here

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on the left hand side of our formula is representing the integral that we're taking here.

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So we have to identify which part of this function here will be you and which part will be DVH.

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Notice that there are no other parts of this integral.

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So we have to assign this entire integral including the dx to you and divi somehow.

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Now for this particular problem I'll go ahead and say that whenever you have the exponential function

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whenever you have e here multiplied by a Trigon the metric function.

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Your best bet will be to assign you to the trick in a metric function and DVH the exponential function.

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So let's go ahead and do that will say that u is equal to cosign of 2 x and that means since it is equal

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to cosign to X that divi here.

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The other part of the integral has to include everything else in our integral which would be either

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the 7 x and the x.

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So we'll say that DV is equal to E to the 7 x dx.

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One thing to note here is that X will always be here with your DVH.

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So now we need to since we have you and DVH we need to take the derivative of you to find D u and we

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need to take the integral of DVM to find the next thing that we have to do.

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Now that we have you and DVH is take the derivative of you to get the view and the integral or the antiderivative

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of DVH defined V.

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So taking the derivative of you we'll call it.

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D'you remember that the derivative of cosign is negative sign.

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And we'll be using chain rule because cosigners our outside function and to X is our inside function.

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So we'll take the derivative of the outside function and get negative sign of 2 x.

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But then chainwheel tells us that we have to multiply by the derivative of the inside function so be

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multiplying that by 2.

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And what we end up with is negative to sign of 2 x this 2 out in front here in the coefficient coming

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as a result of the application of changeroom multiplying by the derivative of 2 x.

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So that's our derivative.

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And of course we need to remember to add the x here whenever we take the derivative.

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Now we take the integral of DV to find v.

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So say that V is equal to when we take the integral of each to the 7 x.

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Remember that we take the the integral of an exponential here.

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We have to divide by the coefficient here on the X we're going to end up with one seventh e m to the

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7 x and that will be our value for V.

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So now that we found each of those four values we can go ahead and plug them into our integration by

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parts formula.

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So we'll say that the integral of our Aboriginal function here.

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So each of the 7 x cosign of 2 x x will be equal to the right hand side of our formula.

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So U-V minus the integral of the d u.

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And we're going to just be grabbing those values straight from the values we calculated here.

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So you at times V when we multiply you and we together we'll get one seventh to the 7 x which is the

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value for V times you cosign of 2 x.

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And I just reordered those terms so that we didn't have to simplify them later.

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So that's U times V minus the integral of v d u.

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So the 172 to the 7 x and D'you hear.

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So notice we have the coefficients negative 2 and positive 1 seventh.

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So when we multiply those together we'll get negative two negative two sevenths.

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So we'll say negative two sevenths times each to the 7 x sine of 2 x the X.

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So that's the application of our integration by parts formula.

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Now we need to go ahead and simplify this so what we'll do is we'll just pull the negative two sevenths

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out in front of our integral.

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And what we'll end up with we have a negative sign here and a negative sign here that will cancel.

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So we'll get positive to sevenths times the integral of each of the 7 x.

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Sign of 2 x x and your first thought might be that we haven't made very much progress.

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But whenever you have an exponential time to train the metric function of using integration by parts

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to take the integral what will often happen is you'll need to end up applying integration by parts twice

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and you'll see how this is going to work out.

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In the end but we're going to have to go ahead and apply integration by parts again and we're going

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to do the same thing we did before where we set you equal to the trick a metric function and divi equal

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to the exponential function.

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So here I will say that you is equal to sign of 2 x and therefore we know that divi must be equal to

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the rest of it which will be easy to the 7 x times the X when we take the derivative of you to get the

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U.

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We know that the derivative of sine is cosign.

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So we'll get cosign of 2 x but remember chain rule tells us that we have to multiply that by the derivative

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of the inside function inside function is to X the derivative of which is 2.

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So our derivative will be to cosign of 2 x times the X and we take the integral of d to find V.

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And again we've already done this.

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We did the same thing up here.

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We know that the integral is one seventh key to the 7 x.

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So now that we've assigned these values we can go ahead and plug in again.

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So what's going to be interesting about this we'll go ahead and rewrite the integral over here on our

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left hand side.

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So cosign to x dx is equal to what we're going to be plugging in for.

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Is the is just the integral above's.

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Is 7 x cosign of 2 x plus 2 7.

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And now the substitution we're making is just for the integral here.

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This whole thing right.

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So we're going to plug in the values of u d u v and DVH just for the integral here.

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Using this formula.

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So we're going to basically make a substitution here like this so we'll say to sevenths and drop in

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parentheses.

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And now we're going to plug in UVM minus the integral of VDU.

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So you at times V will give us one seventh E to the 7 x.

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Sign of 2 x minus the integral of v d u.

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So what we hear at times c you notice we have the coefficient 2 and 1 7.

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So that's a 2 7.

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And we can actually pull that out in front of the integral here as a coefficient on the inside what

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we left with is the rest of the.

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Here you the 7 x and the rest of the D U which is cosigned 2 x 6 so times cosign of 2 x dx and then

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we can go ahead and close our privacy.

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We notice that we have to multiply 2 7 through this whole formula here.

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Now as you can see I just copy the left hand side here and the beginning of the right hand side and

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now we're going to be distributing this to sevenths across both of these values here.

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So we have two seven times one seventh which will give us two over forty nine.

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So two over forty nine times each of the seven x sign of two x and then we have two seven times and

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negative two servants will give us a negative for over forty nine times.

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The integral of each to the seven x cosign of 2 x x and here's why this works out so well.

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What you have to notice now you might be discouraged because you think we ended up right back where

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we started.

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We have even less of the integral of each of the seven x cosign of 2 x x which is exactly the integral

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that we started with so it seems like we've made no progress and we've almost made this more complicated

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when in fact because we have the same integral over here on the right side that we do on the left hand

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side we can go ahead and add this entire negative for 49 times the integral here.

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We can go ahead and add that over the left hand side.

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So imagine you've got four or forty nine here.

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Let's make a common denominator and multiply the left hand side over here.

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I 49 over forty.

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It's the same as multiplying by 1 right.

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And now we can add this entire value to both sides so when we add four forty nine times the integral

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here both sides the value over here on the right hand side will go away.

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This whole value will go away when we add it to both sides and on the left hand side here will be left

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with 49 plus 4 which gives us 53 over forty nine times the integral of each of the seven acts cosign

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of two x dx and on the right hand side we're just left with everything besides the integral.

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Now the last thing we need to do in order to solve for the integral right we're trying to find this

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value here right.

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The integral of either the 7 x cosign of 2 x x and we have that right here.

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So in order to solve for it all we have to do is divide both sides by 53 over 49.

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This coefficient goes away and we've solved four integral.

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So instead of dividing both sides by 53 over 49 we'll go ahead and multiply both sides.

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I 49 over 53 to cancel out what we have here on the left hand side

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what we're left with here is 49 over 53 times.

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1 7.

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And for our second term here

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49 over 53 times to over forty nine either the 7 x signed x

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our final step is to simplify the seven in the denominator here will cancel and we'll just be left with

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seven here in the numerator.

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So we'll get seven over fifty three E to the seven x times cosign of 2 x plus here obviously we'll get

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the 49ers to cancel and we'll be left with two over the three E to the 7 x times sign of 2 x.

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And at this point we can leave our answer this way.

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Or you can choose to factor out one over 53 times each to the 7 x so if we factor this out to try to

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simplify our answer 153 times each of the 7 x what will be left with in our first term here is just

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7 cosign of 2 x plus 2 times sign of 2 x.

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And at this point we hadn't included it earlier because we were doing a lot of finagling and moving

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things around.

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But at this point we have to remember because we did take an integral to go ahead and add plus C to

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account for our constant integration.