We have,
$\int \frac{a \cos x-2 \sin x}{b \sin x+5 \cos x} d x=\frac{7}{41} x+\frac{22}{41}$
$\begin{aligned} & \log |(b \sin x+5 \cos x)|+C \\ & \text { Now, let } a \cos x-2 \sin x=\lambda(b \sin x+5 \cos x) \\ & \qquad+\mu(b \cos x-5 \sin x)\end{aligned}$
$\therefore a=5 \lambda+\mu b$ and $-2=b \lambda-5 \mu$
But it is given, $\lambda=\frac{7}{41}$ and $\mu=\frac{22}{41}$
$a=3, b=4$
$\int \frac{d x}{b+a \cos x}=\int \frac{d x}{4+3 \cos x}$
$=\int \frac{\left(1+\tan ^2 \frac{x}{2}\right) d x}{4\left(1+\tan ^2 \frac{x}{2}\right)+3\left(1-\tan ^2 \frac{x}{2}\right)}$
$\begin{aligned} &=\int \frac{\sec ^2 x / 2}{7+\tan ^2 x / 2} d x \\ & \tan \frac{x}{2}=t \Rightarrow \sec ^2 \frac{x}{2} d x=2 d t \\ & \frac{d x}{b+a \cos x}=2 \int \frac{d t}{7+t^2} \\ &=\frac{2}{\sqrt{7}} \tan ^{-1} \frac{t}{\sqrt{7}}+C \\ &=\frac{2}{\sqrt{7}} \tan ^{-1}\left(\frac{\tan x / 2}{\sqrt{7}}\right)+C\end{aligned}$