Evaluate $\displaystyle \int \dfrac{2 \sin 2 x - \cos x}{6 - \cos^2 x - 4 \sin x} \; dx$
$\begin{aligned}
\text{Let } I & = \int \dfrac{2 \sin 2 x - \cos x}{6 - \cos^2 x - 4 \sin x} \; dx \\\\
& = \int \dfrac{4 \sin x \cos x - \cos x}{6 - \left(1 - \sin^2 x\right) - 4 \sin x} \; dx \;\;\; \left[\text{Note: } \sin 2 x = 2 \sin x \cos x\right] \\\\
& = \int \dfrac{\cos x \left(4 \sin x - 1\right)}{\sin^2 x - 4 \sin x + 5} \; dx \;\;\; \cdots (1)
\end{aligned}$
In equation (1), let $\sin x = t$ $\;\;\; \cdots$ (2a)
Differentiating equation (2a) gives
$\cos x \; dx = dt$ $\;\;\; \cdots$ (2b)
In view of equations (2a) and (2b), equation (1) becomes
$I = \displaystyle \int \dfrac{\left(4 t - 1\right) \; dt}{t^2 - 4 t + 5}$ $\;\;\; \cdots$ (3)
In equation (3), let
$4 t - 1 = P \dfrac{d}{dt} \left(t^2 - 4 t + 5\right) + Q$
i.e. $4 t - 1 = P \left(2 t - 4\right) + Q$ $\;\;\; \cdots$ (4a)
i.e. $4 t - 1 = 2 P t - 4 P + Q$
Comparing the coefficients of t and the constant term on both sides gives
$2 P = 4$ $\implies$ $P = 2$
and $- 1 = -4 P + Q$ $\implies$ $Q = 4 P - 1 = 7$
Substituting the values of P and Q in equation (4a) gives
$4 t - 1 = 2 \left(2 t - 4\right) + 7$ $\;\;\; \cdots$ (4b)
In view of equation (4b), equation (3) becomes
$\begin{aligned}
I & = \int \dfrac{2 \left(2 t - 4\right) + 7}{t^2 - 4 t + 5} \; dt \\\\
& = \int \dfrac{4 t - 8}{t^2 - 4 t + 5} \; dt + 7 \int \dfrac{dt}{t^2 - 4 t + 5} \\\\
& = I_1 + 7 I_2 \;\;\; \cdots (5)
\end{aligned}$
Consider $I_1 = \displaystyle \int \dfrac{4 t - 8}{t^2 - 4 t + 5} \; dt$ $\;\;\; \cdots$ (5a)
In equation (5a), let $t^2 - 4 t + 5 = z$ $\;\;\; \cdots$ (6a)
Differentiating equation (6a) gives
$\left(2 t - 4\right) dt = dz$
$\therefore$ $\left(4 t - 8\right) dt = 2 dz$ $\;\;\; \cdots$ (6b)
In view of equations (6a) and (6b), equation (5a) becomes
$\begin{aligned}
I_1 & = \int \dfrac{2 \; dz}{2} \\\\
& = 2 \; \log \left|z\right| + c_1 \\\\
& = 2 \; \log \left|t^2 - 4 t + 5\right| + c_1 \;\;\; \cdots (7a)
\end{aligned}$
Consider
$\begin{aligned}
I_2 & = \int \dfrac{dt}{t^2 - 4 t + 5} \;\;\; \cdots (5b) \\\\
& = \int \dfrac{dt}{\left(t^2 - 4 t + 4\right) + 5 - 4} \\\\
& = \int \dfrac{dt}{\left(t - 2\right)^2 + \left(1\right)^2} \\\\
& = \tan^{-1} \left(t - 2\right) + c_2 \;\;\; \cdots (7b) \;\;\; \left[\text{Note: } \int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a} \tan^{-1} \left(\dfrac{x}{a}\right) + c\right]
\end{aligned}$
$\therefore$ $\;$ In view of equations (7a) and (7b) equation (5) becomes,
$\begin{aligned}
I & = 2 \log \left|t^2 - 4 t + 5\right| + c_1 + 7 \tan^{-1} \left(t - 2\right) + c_2 \\\\
& = 2 \log \left|\sin^2 x - 4 \sin x + 5\right| + 7 \tan^{-1} \left(\sin x - 2\right) + c \;\;\; \left[\text{from equation (2a)}\right]
\end{aligned}$
where $c = c_1 + c_2$