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Integration by Parts · Acedra
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Integration by Parts
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0:02
so how can we find the indefinite integral of x times e raised to the x to do this we need to use something called integration by parts the integral of u dv is equal to u times v minus the integral of v d u so we need to determine u dv v and u and d u let me put it in a different order
0:41
so which one should we set equal to u and which one should we set equal to dv it's best to set u equal to x because if we do so d u will simply be equal to one times dx and when dealing with constants they're, a lot, easier, to, integrate, than variables so we're going to make dv equal to e to the x dx now v is the integral of dv the integral of e to the x is simply e to the x so now using the formula u times v that's going to be x times e raised to the x minus the integral of v d u v
1:27
is e to the x d u is 1 dx and the integral of e to the x is e to the x plus some constant c so this is the final answer that's the integral of x raised to the or x times e raised to the x let's try another problem let's integrate x times sine x dx so what should we make u and dv equal to so just like before i'm going to make u equal to x so that d u is going to be 1 dx now dv has to be what remains sine x dx now what is the integral of sine the derivative of what function will give us sign
2:23
the integral of sine is negative cosine the derivative of negative cosine is positive sine so using the formula the integral of u dv is equal to u times v minus the integral of v d u so u is x v is negative cosine x and then minus v and d u is 1 dx so far we have negative x cosine x and then these two negative signs will become positive so plus the integral of cosine x dx now the antiderivative of positive cosine is positive sine so the final answer is what you see here negative cosine i
3:13
mean negative x cosine x plus sine x plus c now here's another problem to work on let's integrate x squared ln x so what should we make u and dv equal to in this problem
3:37
if we make u equal to x squared and we make dv equal to ln x dx the integral of ln x will be quite difficult to do unless you know what it is so we're not going to do that so it's best to make dv x squared dx and u is going to be ln x because we know the derivative of l and x the derivative of l and x is 1 over x
4:10
and the antiderivative of x squared the antiderivative of dv will give us v so the antiderivative of x squared dx is x cubed over 3.
4:30
so let's use the formula now
4:38
u times v that's going to be ln x times x cubed over three minus the integral of v times d u and d u is one over x it's supposed to be dx
5:01
so this is equal to 1 3 x cubed times ln x and then we could cancel an x and so we have minus one third the integral of x squared dx
5:25
the integral of x squared is going to be x to the third over three and then plus some constant c so the final answer is one third x cubed ln x minus one over nine x to the third plus c and that concludes this problem
5:48
so let's work on some more examples let's try this problem what is the anti-derivative of the natural log of x how can we use integration by parts to get this answer
6:10
now we can't make u equal well we can't make a dv equal to l and x because we don't know the integral of l and x we're trying to figure that out so therefore we have to make u equal to ln x the derivative of that is one over x dx dv has to be just we can make dv dx if we want to or one dx and so v the integral of one dx is just x so now let's start with the formula and so it's going to be u times v u is ln x and v is x minus the integral of v and then d u is one over x dx so far we have x ln x
7:08
x times one over x the x variables cancel so we have the integral of one dx now the integral of one dx is x so this is the final answer x ln x minus x plus c that is the integral of the natural log of x
7:36
let's find the indefinite integral of x squared sine x dx
7:46
so which part should be u and which part is going to be dv we need to make u equal to x squared and we need to differentiate it twice to bring it down to a constant so this is one of those problems where you have to use integration of you know integration by parts two times d u is going to be 2x dx dv is sine x dx so this is dv and this is going to be u the integral of sine is negative cosine so let's start with the formula
8:30
so u times v that's x squared times negative cosine x minus the integral of v which is negative cosine x times d u that's 2x dx so let's organize what we have negative x squared cosine x and then these two negative signs will become positive so positive the integral of 2x cosine x we could take the 2 and move it to the front so it's just going to be x cosine x now we need to use the integration by parts formula on the integral of x cosine x
9:23
we're going to make u equal to x and dv is going to be cosine x dx the integral of cosine is sine and the derivative of x is 1 dx so using the formula it's going to be u times v which is x times sine x minus the integral of v d u so v is sine x d u is just dx and so the integral of sine is negative cosine and then plus c
10:15
so now what we need to do is replace this part with what we have here so it's going to be plus we need to distribute 2 to everything in here so it's going to be 2x sine x plus this would be plus cosine x but times two so plus two cosine x and we don't need to multiply c by two because c is just some generic constant and this whole thing is equal to the original integral and so this is the answer negative x squared cosine x plus 2x sine x plus 2 cosine x plus c
11:23
now let's work on this problem let's say we have the integral of x squared e to the x dx go ahead and try that problem so let's begin by writing the formula the integral of u dv is equal to u times v minus the integral of v d u now what should we make u equal to and what should be dv
11:56
i prefer making dv e c x dx
12:03
because once we integrate it it's going to stay the same and i want x square to eventually become a constant so this is one of those problems where you need to do use the integration by parts formula twice so we're going to make u equal to x squared
12:22
du is going to be 2x dx so using the formula it's going to be uv that's going to be x squared that's you v is e to the x minus the integral of v which is e to the x times d u that's 2x dx
12:45
so now this becomes
12:49
well before we uh do that let's use integration by parts one more time so i'm going to make u equal to 2x
13:02
du is going to be the derivative of 2x which is 2 but times dx dv we're going to make that equal to e x dx just like we did before and the integral of that is simply e x so we're going to have x squared e x and then minus now let's use the formula again for this part so it's uv so 2x e to the x and then minus the integral of vdu so that's 2 e to the x dx or multiplying those two
13:54
now let's begin by distributing the negative signs so, this, is, going to, be, negative, 2x e to the x and then these two negatives will cancel giving us positive well let's take out the two so positive two integral e to the x dx so our final answer is going to be x squared e c x minus 2 x e to the x anti derivative of e to the x is just e to the x and then plus c now if we want to we can factor out e to the x so we're going to have x squared minus 2x plus 2 and then plus the constant c
14:40
so this right here is our final answer that is the anti-derivative of x squared e to the x what about this problem what is the integral of ln x squared dx try that so for this problem what we need to do is we need to make u equal to ln x squared dv is going to be equal to the other part we're going to make dv equal to dx so v is going to be the integral of dx which is x d u i'm going to write that here i need more space we need to use the chain rule so first let's use the power rule by
15:28
moving the 2 to the front so it's going to be 2 and then we'll keep everything on the inside the same 2 times ln x and then we're going to subtract that by 1 and then take the derivative of the inside the derivative of l and x is one over x so using the formula this is going to be u times v so v is x u is ln x squared minus the integral of v d u v is x d u is 2 ln x times 1 of x and then let's not forget the dx part so we can cancel x and one of x so this becomes x ln x squared
16:20
and then we could take out the two we can move it to the front so minus two integral ln x dx earlier in this video we determined that the integral of l and x is x ln x minus x
16:38
so this is going to be -2 and then we can replace this with x ln x minus x now let's go ahead and distribute the two so our final answer is going to be x times ln x squared minus 2x ln x and then negative 2 times negative x that's going to be positive 2x and then plus c so this is the indefinite integral of ln x squared now what would you do if you were to see a problem like this ln x to the seventh power dx go ahead and try that well first you need to realize that we can rewrite this problem
17:29
a property of logs allows us to move the exponent to the front so make sure you understand this this is not ln x to the seventh power like this if it was we can't move the seven the seven is only affecting the x variable as a result we can move the 7 to the front so this becomes 7 ln x so this is what we're integrating this is equivalent to 7 integral ln x dx and we know the anti-derivative of l and x it's x ln x minus x so this is our final answer it's just 7 x ln x minus 7 x plus c
18:25
now let's try this one let's find the indefinite integral of e raised to the x sine x dx go ahead and work on that
18:41
so for this problem i want to make u equal to sine x
18:50
and then dv i'm going to set that equal to e to the x dx so v is simply going to be e to the x d u is the derivative of sine x which is cosine x so using the formula it's going to be uv that's going to be e sine x minus the integral of v d u so v is e to the x d u is cosine x dx so we need to use integration by parts one more time but this time we're going to make u equal to cosine x instead of sine and dv is going to be the same dv is going to be e to the x dx if v is the same
19:48
d u is going to be the derivative of cosine which is negative sine and then times dx
20:01
so now we're going to have e x sine x minus and then in brackets u v so that's going to be e x cosine x minus the integral of v to u so v is e to the x d u is negative sine x and then dx so let's go ahead and distribute the negative sign i'm going to rewrite the original integral so the integral of e x sine x that's going to be e to the x sine x these two negative signs will become positive and then we need to distribute this negative so it's negative e raised to the x cosine x
20:51
and then we need to apply the negative to this which is going to be negative e to the x sine x now the key to solving this problem is to realize that this term is very similar to the original problem so what we want to do is we want to add this to both sides if we move it to the other side we're going to have 2 e x sine x
21:22
which is equal to this
21:30
now we want to find the value of just one integral e to the x sine x so what we need to do is divide both sides by two so that's we can say that our final answer
21:50
is one half we also factor out e to the x so it's one half e to the x and then times sine x minus cosine x plus c so that's the anti-derivative of e raised to the x times sine x
22:16
now let's move on to our next example let's find the antiderivative of ln x squared divided by x dx so go ahead and try that now the first thing that i want to do is i'm going to rewrite this as ln x squared times 1 over x dx i want to multiply as a product of two functions now what i'm going to do is i'm going to make u equal to ln x squared du will be 2 times ln x to the first power times 1 over x and then dx
23:07
so since i made u ln x squared dv is going to be the other part dv is going to be 1 over x dx so anti-derivative of 1 over x is ln x now using the formula this is going to be u times v u is ln x squared v is just lnx minus the integral of vdu so v is ln x d u is 2 ln x times 1 over x dx so let's put this all together so first i'm going to rewrite the original problem
23:54
that's equal to ln x squared times ln x that's ln x raised to the third power in this 2 i'm going to move to the front here we have ln x times ln x that's ln x squared
24:19
and it's multiplied by one over x which we can move that to the bottom and we need to realize that these two terms are similar so we need to add this to both sides or if we move it to the other side we'll basically add in 1 plus 2 which will become 3. so on the left side we're going to have 3 integral ln x squared over x dx and that's going to equal ln x to the third power so if we divide both sides by 3 this 3 will disappear and we're going to have it underneath here
25:09
and then plus our constant c so this is the answer to the original problem that's how we can find the integral of ln x squared over x now for the sake of practice go ahead and try this finding a definite integral of e to the 3x cosine 4x dx
25:33
so whenever you have an e to the x times the sine or cosine it's good to make the cosine part equal to u that's what i'd like to do personally so in this one i'm going to make u equal to cosine 4x du the derivative of cosine is going to be negative sine 4x and then times the derivative of the angle the derivative of 4x is 4 and then we're going to have times dx so dv in this problem is going to be e to the 3x to dx the antiderivative of e to the 3x it's e to the 3x over 3 which
26:20
we can write as one third e to the 3x so using the formula it's going to be uv so that's 1 3 e to the 3x times cosine 4x minus the integral of v d u so v is one third e to the three x d u that's negative four sine 4x dx
27:01
so what we're going to do now is we're going to move the constants to the front the two negative signs will cancel they will become positive and we can move one third times four to the front so this is going to be plus four over three integral e to the three x sine four x dx now we need to use integration by parts one more time so dv will be the same we don't need to adjust this part of the formula but u is going to be sine 4x as opposed to cosine for x so du is going to change as well
27:51
the derivative of sine is cosine 4x and then times 4 times dx
28:10
so using integration by parts we're going to have u times v so that's one-third
28:19
e to the 3x times sine 4x and then minus the integral of vdu so v is one third e to the three x and then d u that's times four cosine four x d x
28:52
now let's go ahead and distribute four over three so multiply those together that's going to be 4 over 9 e to the 3x sine 4x and then this 2 will also be 4 over 9 but, it's, going, to be, negative four over nine
29:16
actually there's a four here so this is four thirds times four thirds so that's 16 over nine with a negative sign and then we're going to have the integral of e to the 3x cosine 4x dx
29:48
now notice that these two are similar so we're going to add 16 over 9 integral e to the 3x cosine 4xdx to both sides so this is 1 but we can treat it as if it's 9 over 9 just to get common denominators so 9 over 9 plus 16 over 9 that's going to be 25 over 9 integral e to the 3x cosine 4x dx and then that's going to be 1 3 e to the 3x cosine 4x and then plus 4 over 9 e to the 3x sine 4x
30:43
now to get this to 1 what we're going to do is we're going to multiply both sides by the reciprocal of 25 over nine so that is by nine over twenty five so those two fractions will cancel on the left side we're just going to have e to the three x cosine four x dx
31:07
and what i'm going to do i want to keep this in factored form so we have 9 over twenty five from one third and four over nine we can factor out one third and we can also factor out e to the three x
31:32
so here we took out e to the three x we're left with just cosine four x here we took out e to the three x and we took out one third from four over nine so we're gonna have four thirds left over times sine 4x
31:54
the last thing we could do is maybe reduce this so if we divide both numbers by 3 this becomes 3 over 25 e to the 3x and then times cosine 4x plus 4 over 3 sine 4x plus c
32:20
so that's going to be the answer for this problem that's the indefinite integral of e to the 3x cosine 4x
32:51
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