If $\int \frac{sin 2 x}{(3 + 4 cos x)^{3}} d x = \frac{a + b cos x}{c (3 + 4 cos x)^{d}} + e,$ where $e$ is an arbitrary constant and $a, b, c, d$ are positive integers then minimum value of $a + b + c + d$ is equal to

27

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28

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29

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30

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Solution

The correct option is A $29$
$\int \frac{sin 2 x}{(3 + 4 cos x)^{3}} d x = \frac{a + b cos x}{c (3 + 4 cos x)^{d}} + e$ ....(1)

Consider, $I = \int \frac{sin 2 x}{(3 + 4 cos x)^{3}} d x$
Substitute $3 + 4 cos x = t$
$\Rightarrow - 4 sin x d x = d t$

$I = - \frac{1}{8} \int \frac{t - 3}{t^{3}} d t$

$= - \frac{1}{8} \int \frac{1}{t^{2}} - \frac{3}{t^{3}} d t$

$= - \frac{1}{8} (\frac{1}{t} + \frac{3}{2 t^{2}}) + e$

$= \frac{2 t - 3}{16 t^{2}}$

$= \frac{8 cos x + 3}{16 (3 + 4 cos x)^{2}} + e$

$\Rightarrow a = 3, b = 8, c = 16, d = 2$
Hence, $a + b + c + d = 29$

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