Did you follow the hint? What did it give you?
∫_{₀}^{2π }cosh( k a sinθ) sin( k a cosθ) cosθ dθ = πka (how to prove that step by step)???????
where π = pi
∫_{₀}^{2π}= integration from zero to 2π
the book i am reading got mentioned that ; on expanding the cosh and cosine terms, four separate integrals may be obtained. from symmetry it may be shown that three of these integrals are identically zero leaving...
" on expanding the cosh and cosine terms, four separate integrals may be obtained. from symmetry it may be shown that three of these integrals are identically zero leaving " is all i have for the hints.
after i expend,
∫₀2π ((1/2)cosh(a k sinθ)sin(θ+a k cosθ)-(1/2)cosh(a k sinθ)sin(θ-a k cosθ))dθ
this is what i get and don't know how to continue already....
if the question is only this ∫₀2π sin(k a cosθ)cosθdθ ... i still know how to do. with cosh( k a sinθ) multiplying inside I really no idea how to solve... any suggestion ??/