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The value of ...

Question

The value of $\int \frac{cos 4 x + cos 2 x - 6 sin x - 2}{sin 3 x + 2 sin x + 3} d x$ is
(where $C$ is constant of integration)

A

cos 2 x + C

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B

2 cos x + C

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C

2 sin 2 x + sin x + C

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D

2 sin 2 x + C

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Solution

The correct option is B $2 cos x + C$
Let
$I = \int \frac{cos 4 x + cos 2 x - 6 sin x - 2}{sin 3 x + 2 sin x + 3} d x = \int \frac{cos 4 x - cos 2 x + 2 cos 2 x - 2 - 6 sin x}{sin 3 x + 2 sin x + 3} d x = \int \frac{- 2 sin 3 x sin x - 2 (1 - cos 2 x) - 6 sin x}{sin 3 x + 2 sin x + 3} d x = \int \frac{- 2 sin 3 x sin x - 2 (2 {sin}^{2} x) - 6 sin x}{sin 3 x + 2 sin x + 3} d x = \int \frac{- 2 sin x (sin 3 x + 2 sin x + 3)}{sin 3 x + 2 sin x + 3} d x = \int - 2 sin x d x = 2 cos x + C$

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