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Integration as Antiderivative

∫ sin 2xsin 5...

Question

$\int \frac{s i n 2 x}{s i n 5 x s i n 3 x} d x$

A

l n s i n 3 x - l n s i n 5 x + c

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B

\frac{1}{3} l n s i n 3 x + \frac{1}{5} l n s i n 5 x + c

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C

\frac{1}{3} l n s i n 3 x - \frac{1}{5} l n s i n 5 x + c

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D

3 l n s i n 3 x - 5 l n s i n 5 x + c

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Solution

The correct option is C $\frac{1}{3} l n s i n 3 x - \frac{1}{5} l n s i n 5 x + c$
$\int \frac{sin 2 x}{sin 5 x sin 3 x} d x$
$= \int \frac{sin (5 x - 3 x)}{sin 5 x sin 3 x} d x$ replace $2 x = 5 x - 3 x$
$= \int \frac{sin 5 x cos 3 x - cos 5 x sin 3 x}{sin 5 x sin 3 x} d x$ using $sin (A - B) = sin A cos B - cos A sin B$
$= \int \frac{sin 5 x cos 3 x}{sin 5 x sin 3 x} d x - \int \frac{cos 5 x sin 3 x}{sin 5 x sin 3 x} d x$
$= \int \frac{cos 3 x}{sin 3 x} d x - \int \frac{cos 5 x}{sin 5 x} d x$
$= \int cot 3 x d x - \int cot 5 x d x$
$= \frac{1}{3} ln sin 3 x - \frac{1}{5} ln sin 5 x + c$ using the formula $\int cot a x d x = \frac{1}{a} ln sin 3 x$
where $c$ is the constant of integration

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