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Question

dxsin(xa)sin(xb) is equal to
(where ab and C is constant of integration)

A
sin(ba)lnsin(xb)sin(xa)+C
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B
cosec(ba)lnsin(xb)sin(xa)+C
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C
cosec(ab)lnsin(xb)sin(xa)+C
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D
sin(ab)lnsin(xb)sin(xa)+C
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Solution

The correct option is B cosec(ba)lnsin(xb)sin(xa)+C
Let
I=dxsin(xa)sin(xb)
I=1sin(ba)sin((xa)(xb))dxsin(xa)sin(xb)
I=cosec(ba)[cot(xb)dxcot(xa)dx]
I=cosec(ba)[ln|sin(xb)|ln|sin(xa)|]+C
I=cosec(ba)lnsin(xb)sin(xa)+C

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