Question #e4efb

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Question #e4efb

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Answer 1 · 2017-10-12T18:15:38

$\int \frac{sin x}{sin (x + a)} d x = (x + a) cos a - log (sin (x + a)) sin a + C$ $int sinx/sin(x+a) dx = (x+a)cosa-log(sin(x+a))sina + C$

Explanation:

Making $y = x + a$ $y = x+a$ and $d y = d x$ $dy = dx$ we have

$\int \frac{sin x}{sin (x + a)} d x = \int \frac{sin (y - a)}{sin y} d y = \int (cos a - \frac{cos y}{sin y} sin a) d y = y cos a - log (sin y) sin a + C$ $int sinx/sin(x+a) dx = int sin(y-a)/siny dy = int (cosa - cos y/siny sina)dy = y cosa-log(siny) sina + C$

and finally

$\int \frac{sin x}{sin (x + a)} d x = (x + a) cos a - log (sin (x + a)) sin a + C$ $int sinx/sin(x+a) dx = (x+a)cosa-log(sin(x+a))sina + C$

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Question #e4efb

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