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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

`(tanx)/((cosx)^(2005))+C`

C

`-(tanx)/((cosx)^(2005))+C`

D

`(-cotx)/((cosx)^(2005))+C`

Text Solution

Verified by Experts

The correct Answer is:
D
`l=int(cosx)^(-2005)"cosec"^(2)xdx-2005 int(dx)/(cos^(2005)x)`
`l=int(cosx)^(-20005)(-cotx)-int(-2005)(cosx)^(-2006)`
`(-sinx)(-cotx)dx'-2005int(dx)/(cos^(2005)x)`
`l=-(cotx)/((cosx)^(2005))+C`
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