# Cosx cosx integral help

I need help for this integral

cosx cosx$$^{2}$$

gabbagabbahey
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I need help for this integral

cosx cosx$$^{2}$$

Do you mean
$$\int cos(x) cos^2(x) dx$$
or
$$\int cos(x) cos(x^2) dx$$
?

HallsofIvy
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Whatever you mean, what have you tried?

cos(x)cos(x^2)

Defennder
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I don't think the anti-derivative can be expressed in elementary functions.

gabbagabbahey
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cos(x)cos(x^2)

You will need to integrate this by-parts and express your answer in terms of Fresnel Integrals which are defined here: http://http://en.wikipedia.org/wiki/Clothoid#Cornu_spiral" [Broken]

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thanks... sorry for my english...I hope that can understand me...

This integral $$\int_{1}$$ $$^{\infty}$$ cos(x)cos(x2) dx

I have to say that is convergent or divergent. The proffessor gave us a hint (Fresnel Integral)....but I don't know how to use it, He don't discuss it in class :grumpy:. I understand everything but the cos(x2) jeje I don't know how to integrate.:yuck:

gabbagabbahey
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thanks... sorry for my english...I hope that can understand me...

This integral $$\int_{1}$$ $$^{\infty}$$ cos(x)cos(x2) dx

I have to say that is convergent or divergent. The proffessor gave us a hint (Fresnel Integral)....but I don't know how to use it, He don't discuss it in class :grumpy:. I understand everything but the cos(x2) jeje I don't know how to integrate.:yuck:

Okay, this is going to be ugly, but here it goes:

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

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

$$\Rightarrow \int_1^{\infty} cos(x)cos(x^2)dx$$

$$= \frac{1}{2} \int_1^{\infty} cos \left( \left( x - \frac{1}{2} \right) ^2 \right) cos \left( \frac{1}{4} \right) +sin \left( \left( x - \frac{1}{2} \right) ^2 \right) sin \left( \frac{1}{4} \right) +cos \left( \left( x + \frac{1}{2} \right) ^2 \right) cos \left( \frac{1}{4} \right) +sin \left( \left( x + \frac{1}{2} \right) ^2 \right) sin \left( \frac{1}{4} \right) dx$$

$$= \frac{1}{2} \left[ cos \left( \frac{1}{4} \right) \int_1^{\infty} cos \left( \left( x - \frac{1}{2} \right) ^2 \right)dx + sin \left( \frac{1}{4} \right) \int_1^{\infty} sin \left( \left( x - \frac{1}{2} \right) ^2 \right)dx + cos \left( \frac{1}{4} \right) \int_1^{\infty} cos \left( \left( x + \frac{1}{2} \right) ^2 \right) dx$$

$$\left{+ sin \left( \frac{1}{4} \right) \int_1^{\infty} sin \left( \left( x + \frac{1}{2} \right) ^2 \right) dx \right]$$

...

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gabbagabbahey
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...

Make the following substitutions:

$$u \equiv x-\frac{1}{2} , \quad v \equiv x+ \frac{1}{2} \quad \Rightarrow du=dv=dx , \quad u: \frac{3}{2} \rightarrow \infty , \quad v: \frac{1}{2} \rightarrow \infty$$

$$\Rightarrow \int_1^{\infty} cos(x)cos(x^2)dx$$
$$= \frac{1}{2} cos \left( \frac{1}{4} \right) \int_{\frac{3}{2}}^{\infty} cos(u^2)du + \frac{1}{2} sin \left( \frac{1}{4} \right) \int_{\frac{3}{2}}^{\infty} sin(u^2)du + \frac{1}{2} cos \left( \frac{1}{4} \right) \int_{\frac{1}{2}}^{\infty} cos(v^2)dv + \frac{1}{2} sin \left( \frac{1}{4} \right) \int_{\frac{1}{2}}^{\infty} sin(v^2)dv$$

...

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gabbagabbahey
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...
Now using the definition of the Fresnel sine and cosine integrals, $$S(t),\quad C(t)$$ :

$$S(t) \equiv \int_0^t sin(q^2)dq , \quad C(t) \equiv \int_0^t cos(q^2)dq$$

we get:

$$\int_1^{\infty} cos(x)cos(x^2)dx = \frac{1}{2} cos \left( \frac{1}{4} \right) \left( C(\infty)-C \left( \frac{3}{2} \right) \right) + \frac{1}{2} sin \left( \frac{1}{4} \right) \left( S(\infty)-S \left( \frac{3}{2} \right) \right) + \frac{1}{2} cos \left( \frac{1}{4} \right) \left( C(\infty)-C \left( \frac{1}{2} \right) \right)$$
$$+ \frac{1}{2} sin \left( \frac{1}{4} \right) \left( S(\infty)-S \left( \frac{1}{2} \right) \right)$$
....

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gabbagabbahey
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...And using the fact that
$$S(\infty) = C(\infty) =\frac{\sqrt{\pi}}{8}$$
this becomes:

$$\int_1^{\infty} cos(x)cos(x^2)dx = -\frac{1}{2} \left[ cos \left( \frac{1}{4} \right) \left( C \left( \frac{3}{2} \right) + C \left( \frac{1}{2} \right) -\frac{\sqrt{\pi}}{4} \right) + sin \left( \frac{1}{4} \right) \left( S \left( \frac{3}{2} \right) +S \left( \frac{1}{2} \right) -\frac{\sqrt{\pi}}{4} \right) \right]$$

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iah....very ugly...wow!

thanks thanks... jeje now I will try to understand it!!!!!!!!

im nanies roommate and we take the class together and the problem is that i dont understand what u did in the first part of this problem and dont know where the cosX^2 - X come from. also i dont understand why u divided everything by 2...please help me

gabbagabbahey
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im nanies roommate and we take the class together and the problem is that i dont understand what u did in the first part of this problem and dont know where the cosX^2 - X come from. also i dont understand why u divided everything by 2...please help me

I used the following Trig Identity:

$$cos(A)cos(B)=\frac{cos(A-B)+cos(A+B)}{2}$$

ok thank you soo much! the thing is that i have being trying it since tuesday! but thanks

THANK YOU!!!!!!!!!!

a doubt come into my mind when i was doing the calculus exercise. .....i need to know if in the fresnel integral i can used any limit of integration or only from 0 to infinite....

gabbagabbahey
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a doubt come into my mind when i was doing the calculus exercise. .....i need to know if in the fresnel integral i can used any limit of integration or only from 0 to infinite....

If the limits are zero to infinty, then you have S(infinty), but if the limits are 0 to a, you have S(a) as long is a is positive, the fresnel integrals converge.