Question

Integrate 1/sinx+cosx

Answer 1

Answer:

I =Integ.dx/(sinx-cosx). =

Integ.dx/[{2tan x/2/(1+tan^2 x/2)}-{(1-tan^2 x/2)/ (1+tan^2 x/2)]=

Integ.(1+tan^2 x/2)dx/(2tan x/2 - 1 + tan^2 x/2)

=Integ.sec^2 x/2. dx/(tan^2 x/2 + 2tan x/2 - 1)

Let tan x/2 = t => sec^2 x/2 . (1/2) dx = dt

ie sec^2 x/2. dx = 2 dt

I= 2 Integ. dt/(t^2+2t-1)

= 2integ. dt/[(t+1)^2 - 1 - 1]

=2integ. dt/{(t+1)^2 - (√2)^2}

=2 ×(1/2√2) log{(t+1-√2)/t+1+√2)} + c

=(1/√2)log{(tan x/2 + 1 - √2)/(tan x/2 +1 + √2)} +c

Step-by-step explanation:

I hope it will help you....