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intcosec(x-a)cosecxdx=...

`intcosec(x-a)cosecxdx=`

A

`1/(sina)log[sin(x-ax)cosecx]+C`

B

`1/(sina)log[sin(x-a)sinx]+C`

C

`(-1)/(sina)log|sinx cosec(x-a)|+C`

D

`(-1)/(sina)log[sin(x-a)sinx]+C`

Text Solution

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The correct Answer is:
A
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