Integrate Sin(x^1/2): Help Needed

  • Thread starter JonF
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In summary, The conversation is about how to integrate the function \int \sin{(x^{1/2})} dx. The first suggestion is to make the substitution u = \sqrt{x}, followed by using integration by parts. The second suggestion is to use partial fractions. The conversation then shifts to a new integral \int \frac{1}{(4y^{2} - 4y - 3)^{1/2}} dy, with suggestions to complete the square and use a substitution, possibly involving trigonometric functions. The conversation ends with a suggestion to check the integral on a website and a reminder to use the correct syntax when inputting a function.
  • #1
JonF
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1
i for the life of me can't figure out how to integrate this:

[tex] \int \sin{(x^{1/2})} dx [/tex]
 
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  • #2
Make the substitution [tex]u = \sqrt{x}[/tex] which will turn the integral into something like [tex]\int \left(\sin{u} \times 2\sqrt{x}\right) du[/tex]. Apply the substitution again and use integration by parts...
 
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  • #3
We don't have enough x's in there, so use integration by parts to get some. Let u = sin(x^(1/2)) and dv = dx. After that, you'll get an integral that can be solved through a substitution and then a couple of applications of by parts.

cookiemonster
 
  • #4
thanks, got it now
 
  • #5
Ok a new integral I can’t figure out how to take

[tex] \int \frac{1}{4y^{2} - 4y - 3} dy[/tex]

Any suggestions?
 
  • #6
Partial fractions.

cookiemonster
 
  • #7
Oops I posted the wrong integral, I meant:
[tex] \int \frac{1}{(4y^{2} - 4y - 3)^{1/2}} dy[/tex]
 
  • #8
JonF said:
Oops I posted the wrong integral, I meant:
[tex] \int \frac{1}{(4y^{2} - 4y - 3)^{1/2}} dy[/tex]
Try completing the square and using a substitution of what you get inside the ()^2 bit. It should then be a simple matter of knocking it into standard form.
 
  • #9
Try making a change till you get the folowing formula
[tex] \int \frac{1}{((Y-A)^2+B^2)^{1/2}} dY[/tex]
Sorry I forgot the formula of how to go on, you can check it out in your textbooks.
 
  • #10
This might help (a standard form)

[tex]\int \frac{1}{(x^2 + a^2)^{\frac{1}{2}}} dx = \ln \left(x + \sqrt{x^2 + a^2} \right)[/tex]

Though it would appear to me you are on the wrong track. Have checked your integral out at http://integrals.wolfram.com (you need to check their syntax on how to input function it's a little temperamental)
 
  • #11
My suggestion is complete the square on the denominator. Than it should be a matter of using a trig substitution. Looks to me like it will be a secant substitution.
 

1. What is the formula for integrating Sin(x^1/2)?

The formula for integrating Sin(x^1/2) is ∫Sin(x^1/2) dx = 2√x*Sin(2√x)/3 + 2√x*C, where C is the constant of integration.

2. How do I solve ∫Sin(x^1/2) dx?

To solve ∫Sin(x^1/2) dx, you can use the substitution method by letting u = √x. This will change the integral to ∫2Sin(u) du, which can then be solved using the formula for integrating Sin(u).

3. Can I use a trigonometric identity to simplify the integral of Sin(x^1/2)?

Yes, you can use the double angle formula for Sin(2√x) = 2√x*Sin(2√x)/3 to simplify the integral of Sin(x^1/2) to 2√x*Sin(2√x)/3 + 2√x*C.

4. What are the limits of integration for ∫Sin(x^1/2) dx?

The limits of integration for ∫Sin(x^1/2) dx will depend on the specific problem or context in which the integral is being used. In general, the limits should match the variable being integrated with respect to, in this case x.

5. Can I use integration by parts to solve ∫Sin(x^1/2) dx?

Yes, you can use integration by parts to solve ∫Sin(x^1/2) dx by setting u = √x and dv = Sin(x^1/2) dx. However, the substitution method may be a simpler and more efficient approach for this particular integral.

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