Explanation

(a)
Let $I = \int_{}^{}{\frac{\sin{x + 8\cos x}}{4\sin{x + 6\cos x}}dx}$
We can write
$\sin{x + 8\cos x} = A\left( 4\sin x + 6\cos x \right) + B\frac{d}{dx}(4\sin x + 6\cos x)$
$\Rightarrow \sin x + 8\cos x = A\left( 4\sin{x + 6\cos x} \right) + B(4\cos x - 6\sin x)$
On equating the coefficients of $\sin x$ and $\cos x$, we get
$1 = 4A - 6B,\ \ \ 8 = 6A + 4B$
$\Rightarrow \ \ A = 1,\ \ \ \ B = \frac{1}{2}$
$\therefore\ I = \int_{}^{}\frac{\left( 4\sin x + 6\cos x \right) + \frac{1}{2}\left( 4\cos x - 6\sin x \right)}{4\sin{x + 6\cos x}}dx$
$= \int_{}^{}\left( 1 + \frac{1}{2}.\frac{4\cos{x - 6\sin x}}{4\sin{x + 6\cos x}} \right)dx$
$= x + \frac{1}{2}\log{\left( 4\sin x + 6\cos x \right) + c}$