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int(sin(x-a))/(sinx)dx...

`int(sin(x-a))/(sinx)dx`

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Put `(x-alpha)` =t so that `(t+alpha)` and dx=dt.
`:.int(sinx)/(sin(x-alpha))dx=int(sin(t+alpha))/(sint)dt`
`=int(sintcosalpha+costsinalpha)/(sint)dt`
`=cosalpha*intdt+sinalpha*intcottdt`
`=cosalpha*t+sin+alpha*log|sint|+C`
`=(cosalpha)(x-alpha)+sinalpha*log|sin(x-alpha)|+C`.
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