- #1

jmf322

- 18

- 0

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter jmf322
- Start date

- #1

jmf322

- 18

- 0

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

lurflurf

Homework Helper

- 2,453

- 149

sin(x)^2=(1-cos(2x))/2jmf322 said:

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

- #3

- 13,256

- 1,278

Daniel.

- #4

mathmike

- 208

- 0

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

- #5

Jameson

Gold Member

MHB

- 4,538

- 13

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

- #6

toocool_sashi

- 24

- 0

- #7

Jameson

Gold Member

MHB

- 4,538

- 13

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.

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

I believe this is the one.

Last edited by a moderator:

- #8

Orion1

- 973

- 3

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]

Last edited:

- #9

- 13,256

- 1,278

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

Daniel.

Daniel.

- #10

Jameson

Gold Member

MHB

- 4,538

- 13

- #11

mathwonk

Science Advisor

Homework Helper

- 11,391

- 1,626

try parts. that's my favorite way.

- #12

Orion1

- 973

- 3

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/

Last edited:

- #13

mathwonk

Science Advisor

Homework Helper

- 11,391

- 1,626

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:

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:

Share:

- Last Post

- Replies
- 4

- Views
- 344

- Last Post

- Replies
- 8

- Views
- 678

- Replies
- 4

- Views
- 586

- Last Post

- Replies
- 6

- Views
- 649

- Last Post

- Replies
- 3

- Views
- 321

- Replies
- 5

- Views
- 649

- Replies
- 1

- Views
- 385

- Last Post

- Replies
- 1

- Views
- 553

- Last Post

- Replies
- 2

- Views
- 527

- Last Post

- Replies
- 2

- Views
- 461