Question

integral sin x-a/sinx+a
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Answer 1

Answer:

$\large\boxed{\sf{(x + a) \cos(2a) - \sin(2a) log( \sin(x + a) ) }}$

Step-by-step explanation:

$\displaystyle \int \frac{ \sin(x - a) }{ \sin(x + a) } dx$

Let's assume that

$x + a = t \\ \\ = > dx = dt \\ \\ = > x = t - a$

Substituting the values, we get

$= \displaystyle \int \frac{ \sin(t - 2a) }{ \sin(t) }dt$

Further expanding, we get

$= \displaystyle \int \frac{ \sin(t) \cos(2a) - \cos(t) \sin(2a) } { \sin(t) }dt \\ \\ = \cos(2a) \displaystyle \int dt - \sin(2a) \displaystyle \int \cot(t) dt \\ \\ = t \cos(2a ) - \sin(2a) log( \sin(t) )$

Putting the respective values, we get

$\red{ (x + a) \cos(2a) - \sin(2a) log( \sin(x + a) ) }$

Answer 2

Answer:

$\frac{sin \: x - a}{sin \: x + a} \\ \frac{sin \: x - a}{sin \: x + a} \times \frac{sin \: x - a}{sin \: x \: - a} \\ \\ \frac{ {(sin \: x + a)}^{2} }{ {(sin \: x - a)}^{2} } \\ \frac{ {sin }^{2} x+ {a}^{2} + sin \: x \: a}{ {sin}^{2} x + {a}^{2} - sin \: x \: a} \\ answer = 1$

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