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int(cosec^2x-2005)/cos^[2005]x.dx...

`int(cosec^2x-2005)/cos^[2005]x.dx`

A

`(cotx)/((cosx)^(2005))+c`

B

`(tan x)/((cosx)^(2005))+c`

C

`(-(tan x))/((cosx)^(2005))+c`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
D
`int("cosec"^(2)x-2005)/(cos^(2005)x)dx`
`=int(cotx)^(-2005)"cosec"^(2)xdx-2005 int (dx)/(cos^(2005)x)`
`=(cosx)^(-2005)(-cot x)-int (-2005)(cosx)^(-2006)(-sinx)(-cotx)dx-2005int(dx)/(cos^(2005)x)`
`= -(cotx)/((cosx)^(2005))+C`
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