If ∫√1+cosecxdx=f(x)+C, where C is a constant of integration, then f(x) is equal to
A
cos−1(1−2sinx)
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B
sin−1(1−2sinx)
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C
−2sin−1(1−2sinx)
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D
cos−1(1+2sinx)
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Solution
The correct option is Acos−1(1−2sinx) I=∫√1+cosecxdx =∫√sinx+1sinxdx =∫√(1+sinx)(1−sinx)sinx(1−sinx)dx =∫cosx√14−(12−sinx)2dx Put 12−sinx=t Then I=∫−dt√(12)2−t2 =−sin−1(t1/2)+c =−sin−1(1−2sinx)+c =cos−1(1−2sinx)−π2+c =cos−1(1−2sinx)+C where C=c−π2