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Trigonometric Ratios for Sum of Two Angles

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Question

Evaluate: $π \int 0 cos (2 x)$ . $log sin x d x$

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Solution

Solution : -
$I = \int_{0}^{π} c o s 2 x . l o g s i n x d x$

consider $l o g s i n x = I$ $c o s 2 x = I I$

then $I = l o g s i n x \int c o s 2 x d x - \int \frac{d}{d x} (l o g s i n x) . c o s 2 x d x$

$= l o g . s i n . x \frac{s i n 2 x}{2} - \int \frac{c o s x}{s i n x} \times \frac{s i n 2 x}{2} d x$

$= l o g s i n x . \frac{s i n 2 x}{2} - \frac{2}{2} \int c o s^{2} x d x$

$= \frac{1}{2} l o g s i n x . s i n 2 x \frac{- 1}{2} \int (c o s 2 x + 1) d x$

$= [\frac{1}{2} s i n x . s i n 2 x - \frac{1}{2} [\frac{s i n 2 x}{2} + x]]_{0}^{π}$

$= \frac{- π}{2}$

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Trigonometric Ratios for Sum of Two Angles

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