Evaluate int(sqrt(tanx))/(sinx cosx)dx...

Evaluate `int(sqrt(tanx))/(sinx cosx)dx`

Text Solution

Verified by Experts

The correct Answer is:

`2 sqrt(tanx)+C`

`int(sqrt(tanx))/(sinx cosx)dx=int(sqrt(tanx)(1)/("cos"^(2)x))/((sinx cosx)/("cos"^(2)x))dx`
`=int(sqrt(tanx)sec^(2)x)/(tanx)dx=int(tan x)^(-1//2)sec^(2)x dx`
`=((tan x)^(1//2))/(1//2)+C=2 sqrt(tanx)+C`

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The value of int(sqrt(tanx)/(sinx cosx)) dx is equal to

`2sqrt(tanx)+C`

`2sqrt(cotx)+C`

`sqrt(tanx)/(2)+C`

`sqrt(tanx)+C`

Evaluate int((5sinx+6))/(sinx+2cosx+3) dx .

`tan^(-1)((1+tan x/2)/2)+C`

`cot^(-1)((1+"tan" x/2)/2) +C`

`1/2 "tan"^(-1)((1+"tan" x/2)/2)`=+C`

None of these

int(sinx)/(sinx-cosx)dx=

`(1)/(2).log(sinx-cosx)+c`

`(1)/(2).[log(sinx-cosx)+x]+c`

`(1)/(2).log(cosx-sinx)+x+c`

`(1)/(2).[log(cosx-sinx)+x]+c`