Is this Integration of sin^2(x) Correct?

  • Thread starter planck42
  • Start date
  • Tags
    Integration
In summary: Multiply both sides by e^{-ix}, and you get e^{ix}e^{-ix} = (\cos x + \i\sin x)(\cos (-x) + \i\sin (-x)) = \cos^2 x + \sin^2 x = 1. But e^{ix}e^{-ix} = e^{ix - ix} = e^0 = 1. Therefore, \cos^2 x + \sin^2 x = 1, and \cos x and \sin x are related by the Pythagorean identity.
  • #1
planck42
82
0
Could somebody please check my work on the integration of [tex]sin^{2}x dx[/tex]? Thank you for your time.

First step: Complexify the function in order to make straight integration possible

[tex]\sin x=\frac{e^{ix}-e^{-ix}}{2i}[/tex]

[tex]\sin^{2}x=-\frac{1}{2}-\frac{e^{2ix}+e^{-2ix}}{4}[/tex]Second step: Integrate the complex function

[tex]-\frac{1}{4}\int{2+e^{2ix}+e^{-2ix} dx} = -\frac{1}{4}(2x+\frac{e^{2ix}}{2i}-\frac{e^{-2ix}}{2i}+C)[/tex]

[tex]\mbox{However,} \frac{e^{2ix}-e^{-2ix}}{2i} = \sin(2x), \mbox{so}[/tex]
[tex]-\frac{1}{4}(2x+\frac{e^{2ix}}{2i}-\frac{e^{-2ix}}{2i}+C) = -\frac{\frac{1}{2}\sin(2x)+x}{2}+C, \mbox{which appears to be the answer. Can it survive the derivative test?}[/tex]Third step: Take the derivative of [tex]-\frac{\frac{1}{2}\sin(2x)+x}{2}+C[/tex] and see if it equals [tex]\sin^{2}x[/tex]

[tex]\frac{d}{dx}(-\frac{\frac{1}{2}\sin(2x)+x}{2}+C)=-\frac{1}{2}-\frac{1}{2}\cos(2x)=-\frac{1}{2}(1+\cos(2x))=-\cos^{2}x \mbox{, which is the answer less one.}[/tex]

Where is the error in the above process?
 
Physics news on Phys.org
  • #2
I don't think it's that complicated. I would just use the identity [itex]\sin^2x = \frac{1}{2}(1 - \cos2x)[/itex]. that's probably what I would use if I had to reduce (& integrate) any even power of sin or cos, since [itex]\cos^2x = \frac{1}{2}(1 + \cos2x)[/itex]. look at all the other trig identities on wiki:
http://en.wikipedia.org/wiki/Trig_identities
 
Last edited:
  • #3
planck42 said:
Could somebody please check my work on the integration of [tex]sin^{2}x dx[/tex]? Thank you for your time.

First step: Complexify the function in order to make straight integration possible

[tex]\sin x=\frac{e^{ix}-e^{-ix}}{2i}[/tex]

[tex]\sin^{2}x=-\frac{1}{2}-\frac{e^{2ix}+e^{-2ix}}{4}[/tex]
Check this step again. Does this agree with the identity

[tex]\sin^2 x = \frac {1-\cos 2x}2[/tex]
 
  • #4
planck42 said:
Could somebody please check my work on the integration of [tex]sin^{2}x dx[/tex]? Thank you for your time.

First step: Complexify the function in order to make straight integration possible

[tex]\sin x=\frac{e^{ix}-e^{-ix}}{2i}[/tex]

[tex]\sin^{2}x=-\frac{1}{2}-\frac{e^{2ix}+e^{-2ix}}{4}[/tex]
You have a sign error.
[tex]\left(\frac{e^{ix}- e^{-ix}}{2i}\right)^2= -\frac{1}{4}\left({e^{2ix}- 2+ e^{-2ix}\right)[/tex]
[tex]= \frac{1}{2}- \frac{e^{2ix}+ e^{-2ix}}{4}[/tex]


Second step: Integrate the complex function

[tex]-\frac{1}{4}\int{2+e^{2ix}+e^{-2ix} dx} = -\frac{1}{4}(2x+\frac{e^{2ix}}{2i}-\frac{e^{-2ix}}{2i}+C)[/tex]

[tex]\mbox{However,} \frac{e^{2ix}-e^{-2ix}}{2i} = \sin(2x), \mbox{so}[/tex]
[tex]-\frac{1}{4}(2x+\frac{e^{2ix}}{2i}-\frac{e^{-2ix}}{2i}+C) = -\frac{\frac{1}{2}\sin(2x)+x}{2}+C, \mbox{which appears to be the answer. Can it survive the derivative test?}[/tex]


Third step: Take the derivative of [tex]-\frac{\frac{1}{2}\sin(2x)+x}{2}+C[/tex] and see if it equals [tex]\sin^{2}x[/tex]

[tex]\frac{d}{dx}(-\frac{\frac{1}{2}\sin(2x)+x}{2}+C)=-\frac{1}{2}-\frac{1}{2}\cos(2x)=-\frac{1}{2}(1+\cos(2x))=-\cos^{2}x \mbox{, which is the answer less one.}[/tex]

Where is the error in the above process?
 
  • #5
HallsofIvy said:
You have a sign error.
[tex]\left(\frac{e^{ix}- e^{-ix}}{2i}\right)^2= -\frac{1}{4}\left({e^{2ix}- 2+ e^{-2ix}\right)[/tex]
[tex]= \frac{1}{2}- \frac{e^{2ix}+ e^{-2ix}}{4}[/tex]

Wow. What a simple error to overlook. So the final answer is
[tex]-\frac{1}{4}\sin(2x)+\frac{1}{2}x+C[/tex] which does differentiate to [tex]\sin^{2}x[/tex]

Thank you.
 
  • #6
the expansion for sin(x) which you have used is for hyperbolic function of sin(x) .
hyperbolic function of sin(x) is sin(hx).
 
  • #7
No, he has used the correct formula:
[tex]sin(x)= \frac{e^{ix}- e^{-ix}}{2i}[/tex]

The corresponding formula for sinh(x) is
[tex]sinh(x)= \frac{e^x- e^{-x}}{2}[/tex].
 
  • #8
can u explain the formula from where it is derived
 
  • #9
Euler's formula,

[tex]e^{ix} = \cos x + i\sin x[/tex]
 

What is "Integration of sin^2(x)"?

The integration of sin^2(x) refers to the mathematical process of finding the antiderivative of the function sin^2(x).

Why is the integration of sin^2(x) important?

The integration of sin^2(x) is important because it allows us to solve various mathematical problems involving trigonometric functions and their derivatives.

What is the general formula for integrating sin^2(x)?

The general formula for integrating sin^2(x) is ∫sin^2(x) dx = (1/2)x - (1/4)sin(2x) + C, where C is the constant of integration.

What are some common techniques for integrating sin^2(x)?

Some common techniques for integrating sin^2(x) include using the power rule, trigonometric identities, and substitution.

Can the integration of sin^2(x) be applied to real-life situations?

Yes, the integration of sin^2(x) can be applied to real-life situations, such as in physics and engineering, to solve problems involving oscillations, waves, and vibrations.

Similar threads

Replies
6
Views
2K
  • Calculus
Replies
5
Views
1K
Replies
3
Views
304
Replies
3
Views
1K
Replies
16
Views
2K
Replies
4
Views
319
Replies
2
Views
263
Replies
2
Views
908
Back
Top