$\int \frac{2 \sin 2 θ - \cos θ}{6 - \cos^{2} θ - 4 \sin θ} d θ = ?$

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$- 2 \log |\sin^{2} θ - 4 \sin θ + 5| + 7 \tan^{- 1} (\sin θ - 2) + c$

$- 2 \log |\sin^{2} θ - 4 \sin θ + 5| - 7 \tan^{- 1} (\sin θ - 2) + c$

$2 \log |\sin^{2} θ - 4 \sin θ + 5| + 7 \tan^{- 1} (\sin θ - 2) + c$

$2 \log |\sin^{2} θ - 4 \sin θ + 5| - 7 \tan^{- 1} (\sin θ - 2) + c$

answer is A.

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Detailed Solution

$I = \int \frac{2 (2 \sin θ \cos θ) - \cos θ}{6 - (1 - \sin^{2} θ) - 4 \sin θ} d θ \Rightarrow I = \int \frac{(4 \sin θ - 1) \cos θ}{\sin^{2} θ - 4 \sin θ + 5} d θ$

Put $\sin θ = t \Rightarrow \cos θ d θ = d t$

$∴ I = \int \frac{4 t - 1}{t^{2} - 4 t + 5} d t$

Let $4 t - 1 = M \frac{d}{d t} (t^{2} - 4 t + 5) + N$

$\Rightarrow 4 t - 1 = M (2 t - 4) + N$

Comparing the coefficient oft and constant terms on both side, then $M = 2, N = 7$

$\begin{array}{l} ∴ I = \int \frac{2 (2 t - 4) + 7}{t^{2} - 4 t + 5} d t \\ \Rightarrow I = 2 \int \frac{2 t - 4}{t^{2} - 4 t + 5} d t + \int \frac{7 d t}{t^{2} - 4 t + 5} \\ \Rightarrow I = 2 \log |t^{2} - 4 t + 5| + 7 \int \frac{d t}{(t - 2)^{2} + 1} \\ \Rightarrow I = 2 \log |t^{2} - 4 t + 5| + 7 \tan^{- 1} (t - 2) + c \\ \Rightarrow I = 2 \log |\sin^{2} θ - 4 \sin θ + 5| + 7 \tan^{- 1} (\sin θ - 2) + c \end{array}$