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Integrate cos(sin(x))

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Roshdy

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cos(sinx)

can anyone give a result for this integration

∫cos(sin(θ)) dθ

unlimited or limited by any values may reduce any complexity
Roshdy
 

integrate cos(cos x)

I highly doubt that there is a finite expression of that integral. It can be expressed by a series if certain integration limits are applied.
 

    Roshdy

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integration sin(sin(x))

Hi,
If this integral has limits of 0 to pi it will be equal to pi*J(0,x) where J(0,x) is Bessel function of order zero.
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    Roshdy

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integrate cos(cos(x))

cos(sinx)= cos(cos(90-x))=cos^2(90-x)

i.e ∫cos²(90-x)dx which can be ∫ed easily right.
 

integral sin catalan

Hi,
This is not correct because cos(cos(90-x) is not equal to cos^(90-x) whis is equal to cos(90-x)*cos(90-x).
It can only be integrated numerically unless it has limits of 0 to pi where it's Bessel function..
Regards,
 

cos integrates to sin

right
cos(cos(x)) doesn't equal to cos^2(x)
Roshdy
 

What will be the integral
\[ \frac{1}{\pi} \int_0^\pi cos(cos(\theta)) d\theta \]


M
 
Last edited by a moderator:

magnetra said:
What will be the integral
\[ \frac{1}{\pi} \int_0^\pi cos(cos(\theta)) d\theta \]

See previous messages.

\[ \int_{0}^{\pi} \cos ( \sin \tau) \,\mathrm{d}\tau = \int_{0}^{\pi} \cos ( \cos \tau) \,\mathrm{d}\tau\]

--> \[\frac{1}{\pi} \int_{0}^{\pi} \cos ( \cos \tau) \,\mathrm{d}\tau = J_0(1) = 0.7651976865 \]
 
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