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If not stretchy integral fraction numerator sin invisible function application x over denominator sin invisible function application left parenthesis x minus alpha right parenthesis end fraction d x equalsAx + B log sin (x – alpha) + c, then value of (A, B) is –

  1. left parenthesis negative s i n invisible function application alpha comma c o s invisible function application alpha right parenthesis    
  2. left parenthesis c o s invisible function application alpha comma s i n invisible function application alpha right parenthesis    
  3. left parenthesis s i n invisible function application alpha comma c o s invisible function application alpha right parenthesis    
  4. left parenthesis negative c o s invisible function application alpha comma s i n invisible function application alpha right parenthesis    

The correct answer is: left parenthesis c o s invisible function application alpha comma s i n invisible function application alpha right parenthesis


    To find the value of A and B.
    integral fraction numerator sin space x over denominator sin left parenthesis x minus alpha right parenthesis end fraction d x
l e t space x minus alpha equals t
x equals t plus alpha
integral fraction numerator sin left parenthesis t plus alpha right parenthesis over denominator sin space t end fraction d t space equals space integral fraction numerator sin space t space cos space alpha plus sin space alpha space cos space t over denominator sin space t end fraction d t

equals integral cos alpha space d t plus integral sin space alpha space c o t space t space d t

equals cos space alpha left parenthesis x minus alpha right parenthesis plus sin space alpha log ∣ sin left parenthesis x minus alpha right parenthesis ∣ plus c space A
equals cos space alpha space semicolon space B equals sin space alpha
space left parenthesis A comma B right parenthesis equals left parenthesis cos space alpha comma sin space alpha right parenthesis

    Therefore, (A,B)=(cos α,sin α)

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