Product rule in differentiation help

Daniel.I did not know that \cos x=\frac{e^{ix}+e^{-ix}}{2} what is the expression e^{ix} mean?e^{ix} is an exponential function with a complex argument, where i is the imaginary unit. It is a fundamental function in complex analysis and has many useful properties, including the relationship with the trigonometric functions cos(x) and sin(x). It is often used in physics and engineering to simplify calculations involving oscillatory phenomena.
  • #1
UrbanXrisis
1,196
1
[tex] \int x sin (2x) dx =?[/tex]
[tex]u=2x[/tex]
[tex]du=2dx[/tex]
[tex]\int \frac{u}{2} sin(u) \frac{du}{2}[/tex]
[tex]\frac{1}{4} \int usin(u)du[/tex]
[tex]\frac{1}{4} (\frac{u^2}{2}sin(u)+u(-cos(u)))[/tex]
[tex]\frac{2x^2}{4} sin(2x)-\frac{1}{2} cos(2x)+C[/tex]

is this correct?
 
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  • #2
Nope,it's incorrect.U should have part integrated directly

[tex] \int x\sin 2x \ dx=-\frac{1}{2}x\cos 2x+\frac{1}{2}\int \cos 2x \ dx=-\frac{1}{2}x\cos 2x+\frac{1}{4}\sin 2x+\mathcal{C} [/tex]


Daniel.
 
  • #3
Try parts...
 
  • #4
Opps I did not see that you had already posted...
 
  • #5
I haven't learned parts yet, I don't not follow your process
 
  • #6
You can't solve that integral by substitution.
 
  • #7
By parts you set

u = x, du = dx, v = -cos(2x)/2, dv = sin(2x)dx then

[tex] \int u dv = uv - \int v du [/tex] which follows from the product rule of differentiation.

[tex] \int xsin(2x)dx = \frac{-xcos(2x)}{2} + \frac{1}{2}\int{cos(2x)dx} [/tex]

[tex] \int{xsin(2x)dx} = \frac{-1}{2}xcos(2x) + \frac{1}{4}sin(2x)[/tex]
 
  • #8
what exactly is parts and is there a website that explains it?

why can't I solve it with subsitution?
 
  • #9
UrbanXrisis said:
what exactly is parts and is there a website that explains it?

why can't I solve it with subsitution?
Integrating by parts is just taking advantage of the result of the using the product rule in differentiation. See here. You choose a factor to integrate and a factor to differentiate in order to make the integral which appears in the right-hand side of the parts equation easier than the integral you started with.
I doubt you can solve the above by substitution alone, as it involves a product of elementary functions, no matter what substitution you make.
 
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  • #10
so the basic rule to integration by parts is:

[tex]\int u dv= uv- \int v du[/tex]
correct?
 
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  • #11
UrbanXrisis said:
so the basic rule to integration by parts is:

[tex]\int u dv= uv- \int v du[/tex]
correct?
Yep. Choose u and dv such that v*du is a simpler integral. In your case, I would let u = x and v = sin(2x) dx, so that du = dx and v = -(1/2)*cos(2x).
 
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  • #12
U can do it using a substitution & power series,too_Or only power series...

Daniel.
 
  • #13
so I use parts usually when I'm trying to integrate xsin(x) or xcos(x) or xtan(x) ?

what is power series?
 
  • #14
It's quite difficult to integrate

[tex] I=:\int x\tan x \ dx [/tex] (1)

,if u don't know some tricks.

Assume [itex]\cos x\neq 0 [/itex] (2)

Use parts

[tex] \int x\tan x \ dx =-x\ln\cos x+\int \ln\cos x \ dx [/tex] (3)

Denote

[tex] J=:\int \ln\cos x \ dx [/tex] (4)

Now,u know that

[tex] \cos x=\frac{e^{ix}+e^{-ix}}{2} [/tex] (5)

Therefore

[tex] J=\int \ln\left(\frac{e^{ix}+e^{-ix}}{2}\right) \ dx [/tex] (6)

Use the property of the logarithm to write

[tex] J=\int \ln\left[e^{ix}\left(1+e^{-2ix}\right)\right] \ dx -\int \ln 2 \ dx =\int \ln e^{ix} \ dx +\int \ln\left(1+e^{-2ix}\right) \ dx - x\ln 2 [/tex]
[tex]= i\frac{x^{2}}{2}-x\ln 2 + K [/tex](7)

,where

[tex] K=:\int \ln\left(1+e^{-2ix}\right) \ dx [/tex] (8)

Make the substitution

[tex] e^{-2ix}=u [/tex] (9)

[tex] dx=\frac{i}{2u} du [/tex] (10)

Then K becomes

[tex] K=\frac{i}{2}\int \left[\frac{\ln(1+u)}{u}\right] \ du =-\frac{i}{2}\mbox{dilog}(1+u) +\mathcal{C}=-\frac{i}{2}\mbox{dilog}\left(1+e^{-2ix}\right)+\mathcal{C} [/tex] (11)

Therefore,from (1),(3),(4),(7),(11),one gets

[tex] \int x\tan x \ dx=-x\ln\cos x+i\frac{x^{2}}{2}-x\ln 2-\frac{i}{2}\mbox{dilog}\left(1+e^{-2ix}\right)+\mathcal{C} [/tex]

Daniel.
 
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  • #15
UrbanXrisis said:
so I use parts usually when I'm trying to integrate xsin(x) or xcos(x) or xtan(x) ?

what is power series?

Power series:series expansions (Taylor,Laurent) of analytical functions.It's a method to integrate.However,care must be taken with the convergence of such series...


Daniel.
 
  • #16
I did not know that [tex] \cos x=\frac{e^{ix}+e^{-ix}}{2} [/tex]

what is the expression [tex]e^{ix}[/tex] mean?
 
  • #17
The famous Euler's formula

[tex] e^{ix}=\cos x+i\sin x [/tex]


Daniel.
 
  • #18
UrbanXrisis said:
I did not know that [tex] \cos x=\frac{e^{ix}+e^{-ix}}{2} [/tex]

what is the expression [tex]e^{ix}[/tex] mean?

it means, of course, e to the power ix where i is the "imaginary unit": i2= -1.

One problem I have with your posts is that you seem to be posting problems well well in advance of what you know. That is, you post calculus problems that require some basic algebra technique and when someone points that out, you don't know the algebra. I would expect that someone who was taking calculus would have already been exposed to imaginary numbers and, even if not, comfortable with them, have seen eix= cos(x)+ i sin(x) before.
 
  • #19
HallsofIvy said:
One problem I have with your posts is that you seem to be posting problems well well in advance of what you know. That is, you post calculus problems that require some basic algebra technique and when someone points that out, you don't know the algebra. I would expect that someone who was taking calculus would have already been exposed to imaginary numbers and, even if not, comfortable with them, have seen eix= cos(x)+ i sin(x) before.

I guess there is some lack of knowledge on my part then. Probably my basic foundation of math has not been ingrained in my brain yet, that's why I might ask some basic questions to be explained, just to reinforce these foundations. However, I am greatly in debt for all the help and explanations that I have gotten, so thank you all helping me on my 800+ posts :smile:

by the way, I have never seen eix= cos(x)+ i sin(x) before
 
  • #20
In all fairness I've completed the calc requirement for physics and only saw that equation after coming to PF.

[tex] e^{ix}= cos(x)+ i sin(x) [/tex]

Though ofcourse I immediately looked it up and found the derivation and reasoning for it, so I'm cool now. :)
 
  • #21
I haven't been taught that formula in HS maths,but i used in physics when solving problems in alternative current.Instead of phasors and trigonometry,it's better to use complex algebra & trigonometry.

Daniel.
 
  • #22
All you need to know to do the integral you first posted was the method for Integration by Parts, which you seem to grasp now. If you are in Calc I, I wouldn't worry about Euler's identity. You might see some kind of integral where hyperbolic substitution is necessary, but not for this problem. Did you figure out the correct answer?

Jameson
 
  • #23
Jameson said:
All you need to know to do the integral you first posted was the method for Integration by Parts, which you seem to grasp now. If you are in Calc I, I wouldn't worry about Euler's identity. You might see some kind of integral where hyperbolic substitution is necessary, but not for this problem. Did you figure out the correct answer?

Jameson

Yes I understand how to do the question now, my teacher said that he had mistakenly given us that question and that it's a Calc II topic. I did hand in the homework with the Integration by Parts so I think he will be impressed. Also, yes, I am still in high school and for that fact, I will need much guidance to grasp new concepts (such as integration by parts).

This is all leading up to another question. Previously, I was taught the rule that [tex]\int\frac{du}{a^2+u^2} = \frac{1}{a}\arctan\frac{u}{a} + C[/tex]

in a recent example, I saw http://home.earthlink.net/~urban-xrisis/phy001.gif

I'm trying to follow their process but I'm having trouble:
[tex]a=-1[/tex]
[tex]u=-e^{t}[/tex]
[tex]du=-e^{t}[/tex] (du is the derivitive of u right?)

so it would be...
[tex]\frac{1}{-1} tan ^{-1} \frac{-e^t}{-1}[/tex] but I don't know how to incorporate the one minus in the front of the equation.

[tex]\int\frac{du}{a^2+u^2} = \frac{-e^t}{-1^2+(-e^t)^2}[/tex]

what am I doing wrong? how is du = -1?
 
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  • #24
Try the subsititution

[tex] x=a\tan t [/tex]

it will be very simple.

Daniel.
 
  • #25
I'm not catching what you mean by subsitution. Is that a u-subsitution? what does the [tex]_{atant}[/tex] represent? where do I sub that into?
 
  • #26
Into the integral

[tex] \int \frac{dx}{x^{2}+a^{2}} [/tex]

,where else?

Daniel.
 

Related to Product rule in differentiation help

What is the product rule in differentiation?

The product rule is a rule used in calculus to find the derivative of a product of two functions. It states that for two functions, f(x) and g(x), the derivative of their product, f(x)g(x), is equal to f'(x)g(x) + f(x)g'(x).

Why is the product rule important in differentiation?

The product rule is important because it allows us to find the derivative of a product of two functions without having to expand the product and differentiate each term separately. It is a more efficient and straightforward method of finding the derivative.

How do I use the product rule in differentiation?

To use the product rule, first identify the two functions that are being multiplied together. Then, apply the formula f'(x)g(x) + f(x)g'(x) to find the derivative of the product. Remember to take the derivatives of each individual function and plug them into the formula.

Can the product rule be used for more than two functions?

Yes, the product rule can be extended to find the derivative of a product of multiple functions. The general formula for n functions is f'(x)g(x)h(x)... + f(x)g'(x)h(x)... + f(x)g(x)h'(x)... + ... where the pattern continues until the last function has its derivative taken.

Are there any special cases for using the product rule?

Yes, when one of the functions is a constant, its derivative is 0 and can be omitted from the formula. Additionally, when one of the functions is a power function, the power rule can be used instead of the product rule for that specific function.

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