If $integral space fraction numerator sin space straight x over denominator sin space left parenthesis straight x minus straight alpha right parenthesis end fraction space dx space equals space Ax space plus space straight B space log space sin space left parenthesis straight x minus straight alpha right parenthesis space plus straight C comma space space then space value space left parenthesis straight A comma space straight B right parenthesis is$

(sinα, cosα)

(cosα, sinα)

(- sinα, cosα)

(- cosα, sinα)

Solution

(cosα, sinα)

Put x – α = t
$rightwards double arrow space integral fraction numerator sin space left parenthesis straight alpha space plus straight t right parenthesis over denominator sin space straight t end fraction space dt space equals space sin space straight alpha space integral space cot space dt space plus space cos space straight alpha space integral dt space equals space cos space straight alpha space left parenthesis straight x minus straight alpha right parenthesis space plus space sin space straight alpha space ln space vertical line sin space straight t vertical line space plus straight c straight A space equals space cos space straight alpha space comma space straight B space equals space sin space straight alpha$

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