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int(logtanx)/(sinxcosx)dx=?...

`int(logtanx)/(sinxcosx)dx=?`

A

`log{log(tanx)}+C`

B

`(1)/(2)(logtanx)^(2)+C`

C

`log(sinxcosx)+C`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
Put log tan x=t. Then, `(1)/(tanx)*sec^(2)xdx=dt,i.e.,(1)/(sinxcosx)dx=dt`.
`:." "Iinttdt=(1)/(2)t^(2)+C=(1)/(2)(log tanx)^(2)+C`.
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