Question

Evaluate: `int sqrt(tanx)/(sinxcosx) dx`

Answer 1

`I = int sqrt(tanx)/[sinx.cosx]` dx

Dividing numerator and denominator by cosx.

= `int [sqrt(tanx)/cosx]/[(sinxcosx)/(cosx)]` dx

= `int [sqrt(tan x)(1/cosx)]/[(sinx/cosx).cosx]` dx

= `int [sqrt(tan x)]/[sinx/cosx](1/cos^2x)` dx

= `int [sqrt(tan x)]/[tan x](1/cos^2x)` dx

= `int [sqrt(tan x)]/[tan x](sec^2x)` dx

Put, tan x = t
       Sec²x dx = dt

= `int 1/sqrtt dt`

= 2`tan^(1/2) + c`

= 2`sqrttanx` + c    

Answer 2

Given,

`I = int sqrt(tanx)/[sinx.cosx]` dx

 

simplifying the function

 

`= int sqrt(tanx)/[sinx.cosx. (cosx)/(cosx)]` dx

 

`= int sqrt(tanx)/[sinx.(cos^2x)/(cosx)` dx

 

`= int sqrt(tanx)/[cos^2x. (sinx)/(cosx)]` dx

 

`= int sqrt(tanx)/[cos^2x. tanx]` dx

 

`= int [sqrt(tanx).(tan x)^(-1)]/[cos^2x]` dx

 

`= int [(tanx)^(1/2 -1)]/[cos^2x]` dx

 

`= int [(tanx)^(-1/2)]/[cos^2x]` dx

 

`= int (tanx)^(-1/2). 1/[cos^2x]` dx

 

`= int (tanx)^(-1/2). sec^2x` dx

 

Let tan x = t

So, sec²x = `(dt)/(dx)`

 

⇒ dx = `(dt)/(sec^2x)`

 

∴ `int (tanx)^(−1/2).sec^2x` dx

 

= `int(t)^(−1/2).sec^2x. (dt)/(sec^2x)`​

 

= `int(t)^(−1/2)` dt

 

= `(t^(-1/2) + 1)/(-1/2 + 1) + C   ...{as int x^n dx = (x^(n + 1))/(n + 1) + C}`

 

= `t^(1/2)/(1/2) + C`

 

= `2t^(1/2) + C`   

 

= `2sqrt(t) + C`

 

Substituting t = tan x

 

= `2sqrt(tanx) + C`