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int"cosec"(x-a)"cosec"xdx is equal to...

`int"cosec"(x-a)"cosec"xdx` is equal to

A

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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
A
Let `l=int"cosec"(x-a)" cosec" x dx`
`=int(sin a)/(sin a sin(x-a)sinx)dx`
`=-(1)/(sina)int(sin[(x-a)-x])/(sin(x-a)sin)dx`
`=-(1)/(sina) int[(sin(x-a)cosx - cos(x-a)sinx)/(sin(x-a)sinx)]dx`
`=-(1)/(sina)int[cot x - cot(x-a]dx`
`=-(1)/(sina)[log|sinx |-log|sin(x-a)|]+C`
`=-(1)/(sina)[log||sinx" cosec"(x-a)|]+C`
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