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Sum of Trigonometric Ratios in Terms of Their Product

Evaluate: ∫s...

Question

Evaluate:
$\int \frac{sin 2 x d x}{sin 5 x . s i n 3 x}$

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Solution

$\int \frac{sin 2 x}{sin 5 x . sin 3 x} d x$

$sin 2 x = sin (5 x - 3 x)$

$= sin 5 x cos 3 x - cot 5 x sin 3 x$

$\frac{sin 2 x}{sin 5 x . sin 3 x} = \frac{sin 5 x cos 3 x}{sin 5 x sin 3 x} - \frac{cos 5 x sin 3 x}{sin 5 x sin 3 x}$

$= cot 3 x - cot 5 x$

$\int (cot 3 x - cot 5 x) d x$

$= \int cos 3 x d x - \int cot 5 x d x$

$= \int cot 3 x d x = \int \frac{cos 3 x}{sin 3 x} d x = \frac{1}{3} log (sin 3 x) + C_{1}$

$= \int cot 5 x d x = \int \frac{cos 5 x}{sin 5 x} d x = \frac{1}{5} log (sin 5 x) + C_{2}$

$\int (cot 3 x - cot 5 x) d x = \frac{1}{3} log (sin 3 x) - \frac{1}{5} log (sin 5 x) + C$

$\int \frac{sin 2 x}{sin 5 x . sin 3 x} d x = \frac{1}{3} log (sin 3 x) - \frac{1}{5} log (sin 5 x) + C$

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Sum of Trigonometric Ratios in Terms of Their Product

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