Use of derivatives to integrate a function

Question

I want to calculate the integral of $\int \frac {1}{(a+b\cos (x))^2} $ using derivative of a function which partly gives the answer and after manipulations gives the integral. Like for example if we want to calculate the integral of $g (x)=\int \frac {a+b\cos (x)}{(b+a\cos (x))^2} $ we try to develop a function such that $g (x)=\frac {f (x)}{b+a\cos (x)} $ and then differentiate both sides and get $f (x)=\sin (x) $ . Can this way be applied to the integral which I want to calculate?

Answer 1

Note that we can write

$$\int \frac{1}{(a+b\cos(x))^2}\,dx=-\frac{d}{da} \underbrace{\int\frac{1}{a+b\cos(x)}\,dx}_{\text{Evaluate Using Weierstrass Sub}}$$

Answer 2

Answering to your specific question, yo can do $$\frac{a + b\cos x}{(b + a\cos x)^2} = g'(x) = \frac{f'(x)(b + a\cos x) - f(x)(-a\sin x)}{(b + a\cos x)^2},$$ $$a + b\cos x = (b + a\cos x)f'(x) + (a\sin x)f(x).$$ And $f$ will be the solution of a differential equation. Not a good deal.