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int"cosec"^(3)(2x+1)cot(2x+1)dx=?...

`int"cosec"^(3)(2x+1)cot(2x+1)dx=?`

A

`(1)/(4)"cosec"^(4)(2x+1)+C`

B

`-(1)/(3)"cosec"^(3)(2x+1)+C`

C

`-(1)/(6)"cosec"^(3)(2x+1)+C`

D

`(1)/(2)"cosec"(2x+1)cot(2x+1)+C`

Text Solution

Verified by Experts

The correct Answer is:
C
Put `(2x+1)=t anddx=(1)/(2)dt`.
`:." "I=(1)/(2)int("cosec "tcott)dt`.
`-(1)/(2)intu^(2)du", where cosec t"=u`
`=-(1)/(6)u^(3)+C=-(1)/(6)"cosec"^(3)t+C=(-1)/(6)"cosec"^(3)(2x+1)+C`.
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