Assistance with Integration by substitution

In summary, the problem with your equation is that you are missing the minus sign in the first coefficient.
  • #1
12
0

Homework Statement


I have the integral
[tex]\int sin(2.13\sqrt{x}+2.4)\,dx[/tex]

I'm supposed to use the substitution [tex]y=2.13\sqrt{x}+2.4[/tex], aka. [tex]\sqrt{x}=\frac {y-2.4}{2.13}[/tex] to gain the following description of the integral:

[tex]\int sin(2.13\sqrt{x}+2.4)\,dx = E cos(y) + F\int y sin(y)\,dy[/tex]

I have obviously not tried every method possible, but I'm running out of ideas on how to proceed on this one. I tried using the quotient rule without any good looking results, probably due to my non-existing knowledge about these mathematical methods. Any help would be appreciated!
 
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  • #2
Welcome to PF, LizzieL! :smile:

What you need to do is "Integration by substitution".

First you need an expression for x (the square of what you have).
Then you need to differentiate it to find dx.

If you have that you need to replace the parts in your original integral that contain x, by the parts that contain y.
In particular dx needs to be replaced by the derivative you should have for x, followed by "dy".
 
  • #3
Thank you!

Ok, so I have

[tex]x=\big( \frac {y-2.4}{2.13}\big)^2[/tex]

And,
[tex]dx= 2 \big( \frac {y-2.4}{2.13}\big) dy[/tex]

Is this what you mean? I don't think I understood the last part you mentioned though.
 
  • #4
Yes, this is what I meant! :smile:

However, you did not properly apply the chain rule yet.
Do you know what the chain rule is? And more importantly, how to apply it?
As for the last part I mentioned, it was that in your integral you should replace:

[itex](2.13\sqrt{x}+2.4)[/itex] by [itex]y[/itex],

and you should replace:

[itex]dx[/itex] by what you just found (after you correct it for the application of the chain rule).
 
  • #5
Normally I don't think I have a problem understanding the chain rule, but it's all backwards when in integration context, and it confuses me.
My textbook is not covering this topic properly (in my opinion). So to answer your question, I do not know how to apply the chain rule here... :shy:
 
  • #6
Well, in this case it is not backward...
You need to multiply your result for dx with the derivative of [itex](\frac {y-2.4}{2.13})[/itex].

Aaaand... I'm off to bed now! :zzz:
 
  • #7
I'm stuck! :cry:
I'm desperately trying to figure out the derivative of
[tex]y=2.13\sqrt{x}+2.4[/tex]

but I can't seem to get it right.
I'm starting off like this:

[tex]y'=(2.13\sqrt{x}+2.4)'= 2.13x^{1/2} ?[/tex]

How do I calculate this?
 
  • #8
LizzieL said:
I'm stuck! :cry:
I'm desperately trying to figure out the derivative of
[tex]y=2.13\sqrt{x}+2.4[/tex]

but I can't seem to get it right.
I'm starting off like this:

[tex]y'=(2.13\sqrt{x}+2.4)'= 2.13x^{1/2} ?[/tex]

How do I calculate this?
Write your first equation using exponents rather than radicals.
y = 2.13 x1/2 + 2.4

Now, use the power rule to find y'.
 
  • #9
Ok, thanks!
So, is this right or close to right?:

[tex]y^,= 2.13x^{1/2}= 2.13(\frac{1}{2})x^{-1/2}=\frac{2.13}{2\sqrt{x}}[/tex]

And proceeding,

[tex]dy= \frac{2.13}{2\sqrt{x}} dx[/tex] ?

How do I move from here?
 
  • #10
LizzieL said:
Ok, thanks!
So, is this right or close to right?:

[tex]y^,= 2.13x^{1/2}= 2.13(\frac{1}{2})x^{-1/2}=\frac{2.13}{2\sqrt{x}}[/tex]
It's partly right, but you are munging a bunch of stuff together that should be there.

y = 2.13x1.2 + 2.4
so y' = dy/dx = (2.13/2)x-1/2

Don't confuse y with y' - they are different things.

LizzieL said:
And proceeding,

[tex]dy= \frac{2.13}{2\sqrt{x}} dx[/tex] ?

How do I move from here?
Now solve the equation above for dx, which you'll have to replace in your original integral. The original integral looks like this: [itex]\int f(x) dx[/itex]. Use your substitution to replace x and dx with y and dy to get an integral that looks like this: [itex]\int g(y) dy[/itex]. The goal of substitution is to get a different integral that is easier to calculate.
 
  • #11
Thank you for your help!

I think I'm getting there:

[tex]\frac{dy}{dx}= \frac {2.13}{2\sqrt{x}} \Rightarrow
dx = \frac {2\sqrt{x}}{2.13}dy

\Rightarrow
\int sin(y) \cdot \frac {2}{2.13}\Big(\frac {y-2.4}{2.13}\Big)dy[/tex]

[tex]= \frac {-2.4}{2.13^2} cos(y) + \frac {2}{2.13^2}\int y sin(y) dy[/tex]

But I can't seem to get that last part right...the first coeffisient is correct, but not the second one. What did I do wrong?
 
  • #12
Sorry, nevermind.
Your problem is not asking you to do integration by parts.
(Deleted previous post.)
 
  • #13
Your second coefficient is correct.
Why do you think it isn't?

However, your first coefficient is not correct, due to the minus sign.
 

1. What is integration by substitution?

Integration by substitution is a technique used in calculus to evaluate integrals. It involves replacing the variable in the integrand with a new variable, which allows for simplification of the integral.

2. When should I use integration by substitution?

Integration by substitution is typically used when the integrand contains a function within a function, such as f(g(x)). It can also be useful when the integrand contains a complicated expression that can be simplified by substituting a new variable.

3. How do I choose the substitution variable?

The substitution variable should be chosen based on the function within the integrand that is causing difficulty. It should be a variable that allows for simplification of the integrand and makes the integral easier to evaluate.

4. What is the process for integration by substitution?

The process for integration by substitution involves four steps: identifying the function to be substituted, choosing the substitution variable, finding the derivative of the substitution variable, and substituting the variable and its derivative into the integral. This will result in a new integral that can be evaluated more easily.

5. Are there any common mistakes to avoid when using integration by substitution?

One common mistake when using integration by substitution is forgetting to change the limits of integration when substituting the variable. It is important to change the limits to match the new variable. Additionally, it is important to check your work and make sure the answer makes sense in the context of the original problem.

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