What is the integral of #int sqrt(Tan x) / (sin x cos x)dx#?

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Answer 1 · 2018-05-28T14:22:01

The answer is #=2sqrt(tanx)+C#

We need

#tanx=sinx/cosx#

#sinx=cosxtanx=tanx/secx#

Therefore, the integral is

#I=int(sqrt(tanx)dx)/(sinxcosx)=int(sqrt(tanx)dx)/(tanx/secx*1/secx)#

#=int(sec^2xdx)/sqrt(tanx)#

Let #u=tanx#, #=>#, #du=sec^2xdx#

The integral is

#I=int(du)sqrt(u)#

#=sqrt(u)/(1/2)#

#=2sqrt(u)#

#=2sqrt(tanx)+C#