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

1. Jul 6, 2005

jmf322

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

2. Jul 6, 2005

lurflurf

sin(x)^2=(1-cos(2x))/2

3. Jul 7, 2005

dextercioby

If it's $\sin x^{2}$ 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. Jul 8, 2005

mathmike

int of sin ^2 (x) dx = 1/2 x - 1/4 sin 2x + c

5. Jul 8, 2005

Jameson

You're not supposed to give away the answer... tsk tsk

6. Jul 13, 2005

toocool_sashi

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. Jul 13, 2005

Jameson

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.

$$S_1(x) = \sqrt{\frac{{2}}{{\pi}}}\int_{0}^{t}\sin{x^2}dx$$

I believe this is the one.

Last edited: Jul 13, 2005
8. Jul 13, 2005

Orion1

Methodology...

Is this methodology correct?

$$\sin^2 x = \frac{1 - \cos (2x)}{2}$$
$$\int \sin^2 x \; dx = \int \frac{1 - \cos (2x)}{2} dx = \int \frac{1}{2} dx - \int \frac{\cos (2x)}{2} dx$$

$$\int \frac{1}{2} dx - \int \frac{\cos (2x)}{2} dx = \frac{1}{2} \int dx - \frac{1}{2} \int \cos (2x) dx$$

$$\frac{1}{2} \int dx - \frac{1}{2} \int \cos (2x) dx = \frac{1}{2} \left( \int dx - \int \cos (2x) dx \right)$$

$$\frac{1}{2} \left( \int dx - \int \cos (2x) dx \right) = \frac{1}{2} \left( x - \frac{\sin (2x)}{2} \right) + C$$

$$\frac{1}{2} \left( x - \frac{\sin (2x)}{2} \right) + C = \frac{x}{2} - \frac{\sin (2x)}{4} + C$$

$$\boxed{\int \sin^2 x \; dx = \frac{x}{2} - \frac{\sin (2x)}{4} + C}$$

Last edited: Jul 14, 2005
9. Jul 13, 2005

dextercioby

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

Daniel.

10. Jul 13, 2005

Jameson

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. Jul 13, 2005

mathwonk

try parts. thats my favorite way.

12. Jul 14, 2005

Orion1

Methodology...

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

$$\int \sin (x^2) dx = \sqrt{\frac{2}{\pi}} \int_0^t \sin (x^2) dx$$

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

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/

Last edited: Jul 14, 2005
13. Jul 14, 2005

mathwonk

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:

Last edited: Jul 14, 2005