cosh( k a sinθ) sin( k a cosθ) cosθ dθ = πka (how to prove that step by step)???????

where π = pi
= integration from zero to 2​π

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...

Did you follow the hint? What did it give you?

Originally Posted by girdav
Did you follow the hint? What did it give you?
" 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 ??/