Integration by Parts confusions

[O.CModderz]

Hi all ! I'm new here :)

So I'm facing some confusions here regarding integration by parts. While surfing through the internet to study more about this topic, I've came across two formulas which are used in solving problems related to integration by parts.
They are
1. uv - ∫uv'dx
2. uv - ∫vdu

May I know what are the difference between them and how should I use them in solving problems?

Thanks a lot.

 

[Simon Bridge]

Welcome to PF;
Neither of them are complete ... to make sense of them you need to include the LHS:
Your first one should have been... $$\int vu' \;dx = uv - \int uv' \;dx$$... this is integration by parts.

But $u' = du/dx$ so the LHS becomes:$$\int vu' \;dx = \int v\frac{du}{dx} \;dx = \int v\;du$$
Combine with the first one and you get: $$ \int v\;du = uv - \int uv' \;dx \implies \int uv' \;dx = uv-\int v\;du$$... which is your second one.

Putting the two versions in the same form - for comparison:$$\begin{align}\int uv' \;dx &= uv-\int vu'\;dx \\ &= uv-\int v\; du\end{align}$$ ... they are just two different ways of writing the same thing.

 

[O.CModderz]

Thanks for the correction. Basically, either one can be used in problem solving. But I'm still a little confused with how this formula works in a whole. Can you please provide me with a brief explanation on the overall process? For example :

∫sin$^{2}$2x dx

My doubts are :

let u= sin$^{2}$2x and v'=1dx

What happens next with
( u'= ) and ( v= )

well, for now.

Thanks

edit : I used this website to check answers for my questions.
http://symbolab.com/solver/definite-integral-calculator/\int\sin^{2}2xdx
as well as this one
http://www.integral-calculator.com/

 

[SteamKing]

Thanks for the correction. Basically, either one can be used in problem solving. But I'm still a little confused with how this formula works in a whole. Can you please provide me with a brief explanation on the overall process? For example :

∫sin$^{2}$2x dx

My doubts are :

let u= sin$^{2}$2x and v'=1dx

What happens next with
( u'= ) and ( v= )

well, for now.

Thanks

Your post is a little unclear at the end. Are you asking how to take the derivative of sin$^{2}$(2x)? Are you asking how to integrate 1dx?

 

[HallsofIvy]

Thanks for the correction. Basically, either one can be used in problem solving. But I'm still a little confused with how this formula works in a whole. Can you please provide me with a brief explanation on the overall process? For example :

∫sin$^{2}$2x dx

My doubts are :

let u= sin$^{2}$2x and v'=1dx

What happens next with
( u'= ) and ( v= )

You understand that the prime indicates differentiation, don't you? if $u= sin^2(2x)$ the $du= 4sin(2x) cos(2x)$ by the chain rule. And if v'= dx (more correct notation would be "dv= dx") the v= x. So trying to apply "integration by parts" here would give
$$\int u dv= uv- \int vdu$$
$$\int sin^2(x)dx= (sin^2(2x))(x)- 4\int x sin(2x)cos(2x)dx$$

That last integral does not look easier than the last so I would think this integral is NOT a good candidate for "integration by parts"!
(If you really are concerned about integrating this particular function, you would be better advised to use the trig identity $sin^2(\theta)= (1/2)(1- cos(2\theta))$.)

 

[Simon Bridge]

Thanks for the correction. Basically, either one can be used in problem solving. But I'm still a little confused with how this formula works in a whole.

note $(vu)'=vu'+uv'$ (this is just the product rule for differentiation.)
if you integrate both sides you will end up with the integration by parts formula.
The actual method though, is the clever choice of u and v'.

You have to pick the u and v' so that the final integral is easier to solve than the initial one.
It takes practice.

There are plenty of examples online. i.e.
http://tutorial.math.lamar.edu/Classes/CalcII/IntegrationByParts.aspx

 

[O.CModderz]

Your post is a little unclear at the end. Are you asking how to take the derivative of sin$^{2}$(2x)? Are you asking how to integrate 1dx?

I would like to understand how to take the derivative of sin$^{2}$(2x) as what they substituted is 2sin(4x) while what I derived from it was 2cos2xsin2x using chain rule.

 

[O.CModderz]

You understand that the prime indicates differentiation, don't you? if $u= sin^2(2x)$ the $du= 4sin(2x) cos(2x)$ by the chain rule. And if v'= dx (more correct notation would be "dv= dx") the v= x. So trying to apply "integration by parts" here would give
$$\int u dv= uv- \int vdu$$
$$\int sin^2(x)dx= (sin^2(2x))(x)- 4\int x sin(2x)cos(2x)dx$$

That last integral does not look easier than the last so I would think this integral is NOT a good candidate for "integration by parts"!
(If you really are concerned about integrating this particular function, you would be better advised to use the trig identity $sin^2(\theta)= (1/2)(1- cos(2\theta))$.)

Exactly, I know that the prime indicated differentiation. My answer to the differential of u is exactly the same as what you got. However, some of the available online integral calculator got 2sin(4x) instead. This has been bugging me for the past few days.

If its not a good candidate for integration by parts, are there any suggestions for the given example? NOTE : The integral calculator http://symbolab.com/solver/definite-integral-calculator/\int sin^{2}(2x)dx utilized integration by parts.

Thanks a bunch, at least I know that the derivative of sin^2 (2x) that I calculated is correct. :)

 

[Fredrik]

However, some of the available online integral calculator got 2sin(4x) instead.

The identity sin(A+B)=sin A cos B+cos A sin B can be used to rewrite 2sin(4x).

 

[O.CModderz]

I see, thanks for the heads up ! Guess this thread solved all my confusions bout integration by parts. I really appreciate your help!