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

  • Thread starter jmf322
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In summary: Methodology...Yes, your methodology is correct for solving the integral of sin^2(x). As for the intermediate steps, you have successfully integrated by parts and rearranged the resulting integral to get the final answer. The mathematical solution written in symbolic form is:int(sin^2(x)dx) = 1/2 x - 1/4 sin(2x) + C
  • #1
jmf322
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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
 
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  • #2
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
 
  • #3
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.
 
  • #4
int of sin ^2 (x) dx = 1/2 x - 1/4 sin 2x + c
 
  • #5
You're not supposed to give away the answer... tsk tsk
 
  • #6
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.
 
  • #7
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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  • #8
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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  • #9
You should always insert the additive constant in an indefinite integral.

Daniel.
 
  • #10
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.
 
  • #11
try parts. that's my favorite way.
 
  • #12
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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  • #13
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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1. Why does this integral look easy?

Many integrals can appear easy at first glance because they involve simple functions or have a small number of terms. However, the difficulty of an integral depends on many factors, such as the type of function, the limits of integration, and the techniques used to solve it.

2. What makes this integral difficult?

There are several reasons why an integral might not be as easy as it seems. It could involve complex functions, require integration by parts or substitution, or have a tricky set of limits that make it challenging to solve. It's important to carefully analyze the integral and choose the most appropriate method to solve it.

3. Why can't I solve this integral using basic integration rules?

While basic integration rules can be useful, they are not always sufficient to solve every integral. Some integrals require more advanced techniques, such as trigonometric substitution, partial fractions, or integration by parts. It's crucial to understand a variety of integration methods to tackle different types of integrals.

4. Can I use a calculator to solve this integral?

In most cases, calculators can only provide numerical approximations of integrals, not the exact solution. Additionally, integrals involving complex functions or non-standard limits may not be solvable by a calculator. It's important to understand the concepts and techniques of integration to solve integrals accurately and efficiently.

5. Is there a shortcut or trick to solve this integral?

While there may be some shortcuts or tricks for certain types of integrals, it's crucial to have a solid understanding of integration techniques to solve a wide range of integrals accurately. It's always best to carefully analyze the integral and choose the most appropriate method, rather than relying on shortcuts or memorization of specific problems.

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