Evaluate: ∫tanxsinxcosxdx - Mathematics and Statistics

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Sum

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

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Solution 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    

Solution 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`
  Is there an error in this question or solution?
2016-2017 (July)

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