$\int \frac{cos 8 x - cos 7 x}{1 + 2 cos 5 x} d x$ is expressed as $K sin 3 x + M sin 2 x + C$ , then

K = - \frac{1}{3}

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K = \frac{1}{3}

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M = - \frac{1}{2}

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M = \frac{1}{2}

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Solution

The correct options are
B $K = \frac{1}{3}$
C $M = - \frac{1}{2}$
Given : $\int \frac{cos 8 x - cos 7 x}{1 + 2 cos 5 x} d x$

Let $I = \int \frac{cos 8 x - cos 7 x}{1 + 2 cos 5 x} d x$

Considering $\frac{cos 8 x - cos 7 x}{1 + 2 cos 5 x} = \frac{cos 8 x - cos 7 x}{1 + 2 cos 5 x} . \frac{2 sin 5 x}{2 sin 5 x}$
$= \frac{sin 13 x - sin 3 x - sin 12 x + sin 2 x}{2 (sin 5 x + sin 10 x)}$
$= \frac{1}{2} . \frac{sin 13 x + sin 2 x - (sin 3 x + sin 12 x)}{sin 5 x + sin 10 x}$
$= \frac{1}{2} \frac{2 sin \frac{15 x}{2} cos \frac{11 x}{2} - 2 sin \frac{15 x}{2} cos \frac{9 x}{2}}{2 sin \frac{15 x}{2} cos \frac{5 x}{2}}$
$= \frac{1}{2} \frac{cos \frac{11 x}{2} - cos \frac{9 x}{2}}{cos \frac{5 x}{2}} = \frac{- 2 sin 5 x sin \frac{x}{2}}{2 cos \frac{5 x}{2}}$
$= - 2 sin \frac{5 x}{2} sin \frac{x}{2}$

$= cos 3 x - cos 2 x$
Therefore, $\int cos 8 x - cos 7 x = \int cos 3 x - cos 2 x$

$I = \frac{sin 3 x}{3} - \frac{sin 2 x}{2} + c$

Comparing with, $K sin 3 x + M sin 2 x + C$

$K = \frac{1}{3}$ and $M = \frac{- 1}{2}$

Suggest Corrections

Similar questions

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\frac{1}{a} sin a x - \frac{1}{b} sin b x + C

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\int \frac{cos 8 x - cos 7 x}{1 + 2 cos 5 x} d x

\int \frac{cos 7 x - cos 8 x}{1 + 2 cos 5 x} d x

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\frac{1}{k} sin m x - \frac{1}{n} sin p x

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