First slide

Methods of integration

Question

$If I = \int \frac{\sin 2 x}{(3 + 4 \cos x)^{3}} d x, then I equals$

Easy

A

$\frac{3 \cos x + 8}{(3 + 4 \cos x)^{2}} + C$
B

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

$\frac{3 + \cos x}{(3 + 4 \cos x)^{2}} + C$
D

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

Solution

$Write I = \int \frac{2 \sin x \cos x}{(3 + 4 \cos x)^{3}} d x and put 3 + 4 \cos x = t, so that$
$- 4 \sin x d x = d t and$
$I = \frac{- 1}{8} \int \frac{(t - 3)}{t^{3}} d t = \frac{1}{8} (\frac{1}{t} - \frac{3}{2} \frac{1}{t^{2}}) + C$
$= \frac{2 t - 3}{16 t^{2}} = \frac{8 \cos x + 3}{16 (3 + 4 \cos x)^{2}} + C$

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