How Do You Integrate \(\sin(\sqrt{x})\) with Trig Substitution?

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SUMMARY

The discussion focuses on integrating the function \(\sin(\sqrt{x})\) using trigonometric substitution and integration by parts. The recommended substitution is \(u = \sqrt{x}\), which transforms the integral into \(\int \sin(u) \cdot 2\sqrt{x} \, du\). Participants emphasize the necessity of applying integration by parts to derive a solvable integral. Additionally, for the integral \(\int \frac{1}{(4y^{2} - 4y - 3)^{1/2}} dy\), the suggestion is to complete the square and utilize a trigonometric substitution, specifically a secant substitution, to simplify the expression.

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  • Familiarity with integration by parts
  • Knowledge of completing the square in algebra
  • Ability to manipulate integrals involving square roots
NEXT STEPS
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  • Learn about trigonometric substitutions, particularly secant substitution
  • Practice completing the square with various quadratic expressions
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Students and educators in calculus, mathematicians dealing with integral calculus, and anyone seeking to enhance their skills in solving complex integrals through substitution methods.

JonF
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i for the life of me can't figure out how to integrate this:

\int \sin{(x^{1/2})} dx
 
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Make the substitution u = \sqrt{x} which will turn the integral into something like \int \left(\sin{u} \times 2\sqrt{x}\right) du. Apply the substitution again and use integration by parts...
 
Last edited:
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
 
thanks, got it now
 
Ok a new integral I can’t figure out how to take

\int \frac{1}{4y^{2} - 4y - 3} dy

Any suggestions?
 
Partial fractions.

cookiemonster
 
Oops I posted the wrong integral, I meant:
\int \frac{1}{(4y^{2} - 4y - 3)^{1/2}} dy
 
JonF said:
Oops I posted the wrong integral, I meant:
\int \frac{1}{(4y^{2} - 4y - 3)^{1/2}} dy
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.
 
Try making a change till you get the folowing formula
\int \frac{1}{((Y-A)^2+B^2)^{1/2}} dY
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)

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

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.
 

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