Question
Asked Nov 22, 2019
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Evaluate ∫y0ksin(kx−π)dx where y and k are independent of x, and y is given by the solution to the equation cos(ky)=0.57 .

 

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Expert Answer

Step 1

Given:

Jk sin (kx-x
Where y and k are independent of x
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Jk sin (kx-x Where y and k are independent of x

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Step 2

To evaluate the ...

I = [ksin (kx-)dx
= ksin(kx-)dx
cos (kx-)
= k
k
cos (k-
= -cos(ky-)cos(k(0)-z)
=-cos(ky-)cos (-)
Use the following identity cos(-x)cos (x)
(cos(ky) cos(7)+ sin (ky)sin ()+ cos()
(cos(ky) cos +sin (ky )sin z) (-1)
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I = [ksin (kx-)dx = ksin(kx-)dx cos (kx-) = k k cos (k- = -cos(ky-)cos(k(0)-z) =-cos(ky-)cos (-) Use the following identity cos(-x)cos (x) (cos(ky) cos(7)+ sin (ky)sin ()+ cos() (cos(ky) cos +sin (ky )sin z) (-1)

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Tagged in

Math

Calculus

Integration