dirk_mec1 said:So n = 0 here:
[tex] J_n(x) = \frac{1}{\pi} \int_{0}^{\pi} \cos (n \tau - x \sin \tau) \,\mathrm{d}\tau. [/tex]
The formula for integrating cos(sin(x)) from 0 to pi is ∫ cos(sin(x)) dx = −0.5 ∫ (e^x + e^(-x)) sin(x) dx.
To solve the integral of cos(sin(x)) from 0 to pi, you can use the substitution method by letting u = sin(x). This will transform the integral into ∫ cos(u) du which can be easily solved using the trigonometric identity cos(u) = sin(u + π/2).
No, you cannot use the power rule to directly integrate cos(sin(x)) from 0 to pi. This is because cos(sin(x)) is an oscillating function and does not have a constant derivative.
Yes, there is a trick to integrate cos(sin(x)) from 0 to pi. You can use the trigonometric identity cos(sin(x)) = sin(x + π/2) to rewrite the integral as ∫ sin(x + π/2) dx. Then, using the substitution method with u = x + π/2, the integral can be easily solved.
Specifying the limits of integration is important because it helps to determine the exact area under the curve of cos(sin(x)) from 0 to pi. Without specifying the limits, the integral would result in an indefinite integral with a constant of integration, making it impossible to find the exact value.