This looks like an easy integral, why cant i get it

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the integral of sin(x)^2

I tried using all kinds of trigonometric identities, but they never seemed to make it easier. Any suggestions on what identities to use, or how to approach this? Thanks
 

lurflurf

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jmf322 said:
the integral of sin(x)^2

I tried using all kinds of trigonometric identities, but they never seemed to make it easier. Any suggestions on what identities to use, or how to approach this? Thanks
sin(x)^2=(1-cos(2x))/2
 

dextercioby

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If it's [itex] \sin x^{2} [/itex] the function you wish to antidifferentiate, you can express the result in terms of a Fresnel integral, "S" if i'm not mistaking.

Daniel.
 
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int of sin ^2 (x) dx = 1/2 x - 1/4 sin 2x + c
 
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You're not supposed to give away the answer... tsk tsk
 
ru talking of sin(x)sin(x) or is it sin (x*x)??? the 1st one seems quite easy...1-cos(2x) / 2 wud do it....for the second one...integrate by parts taking x^0 as the second function...then u will get sumthing like integral of x^2cos(x^2) in the second integral...integrate partially again...u will get LHS with a minus sign on the RHS(check it out...im not very sure...i did it in rough ;)) ... bring it over to LHS and finish it...check it out...im not so sure of my arithmetic.
 
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I'm fairly sure that the second interpretation cannot be integrated by parts. As Daniel said, it requires the Fresnel integral, which integrals.com confirmed.

[tex]S_1(x) = \sqrt{\frac{{2}}{{\pi}}}\int_{0}^{t}\sin{x^2}dx[/tex]

I believe this is the one.
 
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Methodology...



Is this methodology correct?

[tex]\sin^2 x = \frac{1 - \cos (2x)}{2}[/tex]
[tex]\int \sin^2 x \; dx = \int \frac{1 - \cos (2x)}{2} dx = \int \frac{1}{2} dx - \int \frac{\cos (2x)}{2} dx[/tex]

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

[tex]\frac{1}{2} \int dx - \frac{1}{2} \int \cos (2x) dx = \frac{1}{2} \left( \int dx - \int \cos (2x) dx \right)[/tex]

[tex]\frac{1}{2} \left( \int dx - \int \cos (2x) dx \right) = \frac{1}{2} \left( x - \frac{\sin (2x)}{2} \right) + C[/tex]

[tex]\frac{1}{2} \left( x - \frac{\sin (2x)}{2} \right) + C = \frac{x}{2} - \frac{\sin (2x)}{4} + C[/tex]

[tex]\boxed{\int \sin^2 x \; dx = \frac{x}{2} - \frac{\sin (2x)}{4} + C}[/tex]
 
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dextercioby

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You should always insert the additive constant in an indefinite integral.

Daniel.
 
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Yes, the "+ C" is needed, but besides that you are correct. And you had a correct approach. That's a common Calc I question I believe.
 

mathwonk

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try parts. thats my favorite way.
 
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Methodology...



Mathematica Fresnel Integral definition:
[tex]S(z) = \int_0^z \sin \left( \frac{\pi t^2}{2} \right) dt[/tex]

[tex]\int \sin (x^2) dx = \sqrt{\frac{2}{\pi}} \int_0^t \sin (x^2) dx[/tex]

integrals.wolfram.com solution:
[tex]\boxed{\int \sin (x^2) \; dx = \sqrt{\frac{\pi}{2}} \cdot \text{FresnelS} \left[ \sqrt{\frac{2}{\pi}} x \right]}[/tex]

Is this methodology correct? and what intermediate steps are missing?

What is the integrals.wolfram.com mathematical solution written in symbolic form?


Reference:
https://www.physicsforums.com/showpost.php?p=676949&postcount=8
http://integrals.wolfram.com/
 
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mathwonk

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for the integral sin^2(x) parts gives

int(sin^2(x)dx) = -cos(x)sin(x) + int(cos^2(x)dx)

= -cos(x)sin(x) + int(1 - sin^2(x)dx)

so 2 int(sin^2(x)dx) = x - cos(x) sin(x) +constant.

this way you do not need to have any tricky formulas at hand like

sin2 = (1/2)(1-cos(2x)), and the answer comes slightly simpler too.

sin(x^2) on the other hand is no easier than the famous e^(x^2).

intuition should suggest (by the product rule) that most functions of form

f(g(x)) that cannot be expanded or simplified are not going to occur as elementary derivatives.

with a bit of work one can do sqrt(1+x^2) but only because the substitution x = tan(u) simplifies it.

one cannot do sqrt(1+x^4) this way though. :rofl:
 
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