yes it is true

Thank you for answering. I find my mistake.

Using given change of variables  z= z(x),  you can compute   (1+z^2)^(-3⁄2)dz.=

(1+z(x)^2)^(-3⁄2) z'(x) dx = f(x) dx.

Hint:

Let $A= a +b \cos (x-y)$ and $B=a +b \cos (x+y)$.

Find $AB$ ?

$AB= a^2 + ab ( \cos (x-y) + \cos (x+y) ) +b^2 \cos (x-y) \cos (x+y) $

Since $\cos (x-y) + \cos (x+y)= 2 \cos x \cos y$ and $ \cos (x-y) \cos (x+y)= (\cos 2x + \cos 2y)/2= \cos^2x- \sin^2y$, we find

$AB= a^2 + 2 ab \cos x \cos y + b^2 \cos^2x - b^2 \sin^2y $. Set $t= b \cos x + a \cos y $, we find $AB= t^2 + ( a^2- b^2 )\sin^2y $.

The proof is correct  as Muhammad Nazim Butt  says.

Dear Materjevic,

Thank you for answering kingly.

Sincerely,

Susumu Osawa