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int(sqrt(tanx)+sqrt(cotx))dx is equal to...

`int(sqrt(tanx)+sqrt(cotx))dx` is equal to

A

`sqrt2 tan^(-1)((tanx)/(sqrt(2tanx)))+C`

B

`sqrt2 tan^(-1)((tanx-1)/(sqrt(2tanx)))+C`

C

`(tanx)/(sqrt2).tan^(-1)((cotx+1)/(sqrt(2tanx)))+C`

D

`(tanx)/(sqrt2).tan^(-1)((cotx+1)/(sqrt(2tanx)))+C`

Text Solution

Verified by Experts

The correct Answer is:
B
Let `l=int((sin x+cosx))/(sqrt(sinx.cosx))dx`
`=int(sqrt2(sinx+cosx))/(sqrt(2sinx.cosx))dx=sqrt2 int(sinx+cosx)/(sqrt(sin2x))dx`
`"Put "sinx-cosx=t`
`rArr" "(cosx+sinx)dx=dt`
`"Also, "sin 2x=(1-t^(2))`
`therefore" "l=sqrt2 int(dt)/(sqrt(1-t^(2)))=sqrt2 sin^(-1)t+C`
`=sqrt2 sin^(-1)(sinx - cosx)+C`
`=sqrt2 sin^(-1)(sinx-cosx)+C`
`=sqrt2 tan^(-1)((tanx-1)/(sqrt(2tanx)))+C`
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