Find the following integral: int frac{sin x}{ | vedclass

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(A) Let $x+a = t$. Then $dx = dt$ and $x = t-a$.Substituting these into the integral:$\int \frac{\sin x}{\sin (x+a)} dx = \int \frac{\sin (t-a)}{\sin

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$x \cos a - \sin a \log |\sin (x+a)| + C$

Vedclass · Answer

(A) Let $x+a = t$. Then $dx = dt$ and $x = t-a$.Substituting these into the integral:$\int \frac{\sin x}{\sin (x+a)} dx = \int \frac{\sin (t-a)}{\sin t} dt$Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$:$= \int \frac{\sin t \cos a - \cos t \sin a}{\sin t} dt$$= \int (\cos a - \cot t \sin a) dt$$= \cos a \int dt - \sin a \int \cot t dt$$= t \cos a - \sin a \log |\sin t| + C$Substituting $t = x+a$ back:$= (x+a) \cos a - \sin a \log |\sin (x+a)| + C$$= x \cos a + a \cos a - \sin a \log |\sin (x+a)| + C$Since $a \cos a$ is a constant,we can absorb it into the arbitrary constant $C$:$= x \cos a - \sin a \log |\sin (x+a)| + C$

Find the following integral: $\int \frac{\sin x}{\sin (x+a)} d x$

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