<s>[INST] differentiation under integral sign How do you find:

$$\int_0^\pi\frac{dx}{(5+3\cos x)^3}$$
and
$$\int_0^\pi\frac{\sin^2x }{(5+3\cos x)^3}dx$$
by differentiating under the integral sign?
I know you can get the result by using $\tan\left(\frac{x}{2}\right)$ substitution but I need to solve it via Feynman's Trick. 
I am given the general result that:
$$\int_0^\pi\frac{dx}{a+b\cos x}= \frac{\pi}{\sqrt{a^2-b^2}}$$
 [/INST] $f(a) = \int_0^\pi\frac{dx}{(a+b\cos x)} = \frac {\pi}{\sqrt {a^2 - b^2}}\\
\frac {df}{da} = \int_0^\pi-(a+b\cos x)^{-2} dx = -a\pi(a^2 - b^2)^{-\frac 32}\\
\frac {d^2f}{da^2} = \int_0^\pi 2(a+b\cos x)^{-3}dx = \pi(-(a^2 - b^2)^{-\frac 32}+3a^2(a^2 - b^2)^{-\frac 52})\\
\int_0^\pi (a+b\cos x)^{-3}dx = \frac \pi2 (-(a^2 - b^2)^{-\frac 32}+3a^2(a^2 - b^2)^{-\frac 52})$
$\\$
$g(b) = \int_0^\pi (a+b\cos x)^{-1} dx = \pi(a^2 - b^2)^{-\frac 12}\\
g'(b) = \int_0^\pi -(\cos x)(a+b\cos x)^{-2} dx = b\pi(a^2 - b^2)^{-\frac 32}\\
\frac 12 g''(b) = \int_0^\pi (\cos^2 x)(a+b\cos x)^{-3} dx = \frac \pi2 ((a^2 - b^2)^{-\frac 32}+3b^2(a^2 - b^2)^{-\frac 52})\\
\int_0^\pi (\sin^2 x)(a+b\cos x)^{-3} dx = \frac 12(f''(a) - g''(b))$
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