How does integration by parts work?

[Tino]

Integration by parts HELP !

Ok to be honest with all of you reading this post, i just don't understand how integration by parts work.
Can someone please explain how it works?
I have looked on the internet for help reading through all the notes but i still do not understand.
So please someone explain to me in the easiest way.

Here is an example :

Integrate (2x Sin 3x) dx

 

[arildno]

1. Do you understand the product rule of differentiation?
Essentially, integration by parts is nothing but the reverse operation of the product rule.

2. Let us construct a function h(x)=u(x)v(x)
Then, the product rule gives us:
\frac{dh}{dx}=\frac{du}{dx}v(x)+u(x)\frac{dv}{dx}
This can of course be rewritten as:
u(x)\frac{dv}{dx}=\frac{dh}{dx}-\frac{du}{dx}v(x)
This is the form of the product rule we use in integration by parts.

3. Integration by parts.
Suppose we may write an integrand in the form of u(x)\frac{dv}{dx}
(in your particular example, we may choose, say u(x)=2x, \frac{dv}{dx}=\sin(3x))
Then, from 2, we have:
\int{u}(x)\frac{dv}{dx}dx=\int\frac{dh}{dx}dx-\int\frac{du}{dx}v(x)dx

Are you following thus far?

 

[steven187]

hello there

the only time you would use integration by parts is when you have the product of 2 functions one in which you can reduce which is the u(x)=2x through differentiation so that it equals to a constant or you would want to reduce it to something that will get eliminated once multiplied by the integral of \frac{dv}{dx} so that it gets reduced to something that is integrable for example try doing integration by parts for

\int \log x dx

if you can do this example and the one you are having trouble with you would pretty much prepared to use integration by parts when necessary

 

[Muzza]

I see, so one wouldn't want to use integration by parts to integrate sin(x) * e^x, then ;)

 

[steven187]

I see, so one wouldn't want to use integration by parts to integrate sin(x) * e^x, then ;)

woops i forgot one more sanario thanxs muzza lucky you mentioned it, a person tryin to do that for the first time would probably end up in circles lol that's pretty complicated to explain unless you actually do it,

 

[Jameson]

I see, so one wouldn't want to use integration by parts to integrate sin(x) * e^x, then ;)

That integral requires integration by parts twice and then some rearranging, but it can be solved by this method.

 

[Jameson]

As for the original integral...

\int2x\sin{(3x)}dx

dv = sin(3x)
u = 2x

\int udv = uv - \int vdu

\int2x\sin{(3x)}dx = 2x\int sin{(3x)}dx - \int 6\cos{(3x)}dx

 

[Muzza]

That integral requires integration by parts twice and then some rearranging, but it can be solved by this method.

Yes, precisely my point...

 

[Tino]

Hmmm ??

Ok thanks for all your reply but for the problem i have, so far i understand the rule :

integrate u dv = u v - integrate v du

i have put all the bits into the equation from this problem and it looks something like this, i don't know if its right.

integrate (2x)(Sin 3x dx) = 2x (-1/3 Cos 3x) - integrate (-1/3 Cos 3x)(2 dx)

now here is the problem what do i do after this ?

Please help

Thanks a bunch !

 

[cronxeh]

Continue until you solve it (meaning you might have to do integration by parts or integration by substitution again and again and again and again until everything is killed off)

In case you succeed make sure to check for correct answer:

\frac{2}{9} sin(3x) - \frac{2}{3}x cos(3x) + C

 

[Tino]

hmm don't understand lol

Ok i get what u mean by integrating the equation but, how do i do it :

can u show me the steps because i mean there are two integration i have to do but how do i but them together ?

please show the steps and explain if you can !

 

[steven187]

hello tin0

well all you have to do is evaluate the right hand side and this is done by integrating the second term, you do know how to integrate cos3x coz really that's all you have to get your answer, you have already integrated sin3x so you should be able to do this integration, anyway i hope i have given you more than enough hints to get you answer

 

[cronxeh]

integrate (2x)(Sin 3x dx) = 2x (-1/3 Cos 3x) - integrate (-1/3 Cos 3x)(2 dx)

As you've said it yourself your problem now simplified to
2x (-1/3 Cos 3x) - integrate (-1/3 Cos 3x)(2 dx)

Now go ahead and find \int{\frac{-2}{3} cos(3x) dx using substitution method (try substituting something in cosine to make it a very easy trig function that can be easily integrated, and don't forget to check for any additional things you would need after substituion (like perhaps checking if your new substituted's variable derivative has enough numbers for that original integral)

 

[Tino]

I tried

So far i got to is :

integrate (2x Sin 3x) dx

u = 2x
dv = Sin 3x dx
du/dx = 2
du = 2 dx
v = -1/3 Cos 3x

integrate (2x)(Sin 3x dx) = 2x (-1/3 Cos 3x) - integrate (-1/3 Cos 3x)(2 dx)
= 2x (-1/3 Cos 3x) - integrate -2/3 Cos (3x) dx

But i have never even learned how to use substitution by trigometry method so is there any other way i can do it ?

if i do integration by substitution with this :

integrate -2/3 Cos (3x) dx

u = -2/3 Cos (3x) or u = -2/3 Cos

du/dx = 2/3 3Sin 3x or du/dx = 2/3 Sin

I don't know what to do this is really confusing lol ?

 

[steven187]

hello there

well let's use another dummy variable r since we have already used u anyway
let r=3x
dr/dx=3
dx=dr/3 then use the substitution method .
now the most easiest way to do it is to look up a full table of standard integals and look for something that looks like what you are working with, it isn't that complicated

steven

 

[Tino]

I am still confused !

Ok integration by substitution :

integrate -2/3 Cos (3x) dx

r = -2/3 cos 3x why r = 3x ?

lol

can u show me the steps how to do it ?

 

[cronxeh]

You can take out the coefficients outside because they are not to be integrated - they are just coefficients without the 'x' so to speak. The simplest trig function for you to do is cos(u). So you have 2/3 integral ( cos(3x) dx). Now what do you think you should do to make it into 2/3 integral (cos(u) du)? You substitute:

u = 3x
du = 3dx

1/3 * -2/3 integrate (cos(u) du)

Since you need 3dx in order to substitute for u you have to multiply 3 by cos(u), but now you must balance it out by multiplying by 1/3 outside. Hence your answer ends up being 2/9 sin(3x) for this integral. Add it up with previous results and your answer is

2/9 sin(3x) - 2/3 x*cos(3x) + C

 

[Tino]

woohoo

Thanks a lot people i actucally sort of understand now, thanks a lot !

Thanks peeps !

 

[Tino]

One more little question !

The question was:
integrate (2x Sin 3x) dx

and i put this into :

integrate (2x)(Sin 3x dx) = 2x (-1/3 Cos 3x) - integrate (-1/3 Cos 3x)(2 dx)

i have asked my teacher and he told me to leave 2x (-1/3 Cos 3x) and not change it would that be right ?? or do i change it to :

2/9 sin (3x)

?

 

[steven187]

hello there

this is the answer
\frac{2}{9} sin(3x) - \frac{2}{3}x cos(3x) + C
there is nothing to change, listen to your teacher, especially if your teacher told you not to change anything why do you want to change it? if you want to change the order of the terms around in the answer feel free to do so because that won't change the answer anyway good luck with it

steven

 

[Tino]

Another question lol

Sorry to bother you people but i have another question i got to ask if i have done it right, this time is integration by substitution:

integrate 1 / (3x+5) to the power of 4 dx

u = 3x + 5
du/dx = 3
dx = du / 3

integrate 1 / (u to the power of 4) (du / 3)

1/3 integrate 1 / u to the power of 4 du

1/3 integrate u to the power of -4 du

1/3 integrate u to the power of -3 / u to the power of -3 + C

-1/9 1 / u to the power of 3 + C

-1/9 1 / (3x + 5) to the power of 4 + C

Is this right ?

Thanks

 

[steven187]

hello Tino

Oh your not bothering us, no need to apologise we are here to help each other and we are happy to have you as part of our forum :) now let us look at your question

steven

 

[Tino]

Thanks

Thanks, sorry i will learn to type in different format next time so it will be easier for all of you to read and understand.

 

[steven187]

this is how i would do it

integrate 1/(3x+5)^4 dx
let u = 3x+5
du/dx=3
du=du/3
dx=du/3 now substitute this and u = 3x+5 into the integration question to get
1/3 integrate 1/u^4 du
1/3*-1/3*1/u^3+C=-1/9*1/(3x+5)^3+C=-1/9(3x+5)^3+C
watch very carefully how i have done it

steven

 

[cronxeh]

\int \frac{1}{(3x+5)^4} dx = \frac{-1}{9(3x+5)^3}

 

[Tino]

Thanks for helping !

But was my answer totally wrong -\frac{1}{9} \frac{1}{(3x+5)^3} + C ?

or was it the same as \frac{-1}{9(3x+5)^3} + C ?

where did i go wrong ?

 

[cronxeh]

Your answer is correct but I'm used to simply combining the numerators - and you can put the minus before the fraction or on top or on the bottom of it - as long as you don't forget to have your negative sign there.

 

[Tino]

Thanks

Thanks for your help anyways !

 

[Tino]

Another silly question

This is back to the integration by parts question :

\int (2x Sin 3x) dx

when i put all the u, v, dx etc into the equation below:

\int udv = uv - \int vdu

It looks something like this right:

\int2x\sin{(3x)}dx = (2x)(-\frac{1}{3} cos{(3x)}+c - \int (-\frac{1}{3} cos 3x) 2 dx

then i integrate the last one right and leave everything the same as it is.

\int2x\sin{(3x)}dx = (2x)(-\frac{1}{3} cos{(3x)} - \frac{2}{3} x cos(3x)+C

But i figured out that the cos at the end should be sin ? shouldn't it after integration ?

 

[mattmns]

Yes the last part should be - \int (-\frac{1}{3} cos 3x) 2 dx = \frac{2}{9}sin(3x) + C