Question

integration of 1-tanx/1+tanx

Answer 1

(1-tan x)/(1+tan x)=(1-(sin x)/(cos x ))/(1+(sin x)/(cos x)) = (cos x - sin x )/(cos x + sin x ) = (cos x cos (pi/4)-sin x sin (pi/4))/(cos x cos (pi/4) + sin x sin (pi/4) )=cot(x+pi/4)#
Note that #sin(pi/4)=cosw(pi/4)=1/sqrt 2

8

Answer 2

Answer:

Integration of (1-tanx)/(1+tanx) = ln(cosx + sinx) + C

Step-by-step explanation:

$\int\limits {\frac{1-tanx}{1+tanx} } \, dx = \int\limits {\frac{cosx-sinx}{cosx+sinx} } \, dx$  ( Put $tanx = \frac{sinx}{cosx}$)-----------(1)

Let,  $cosx + sinx = t$

Differentiating the both sides,

$(-sinx + cosx) dx = dt$

⇒$(cosx - sinx) dx = dt$----------------(2)

From equation (1) & (2),

$\int\limits {\frac{cox - sinx}{cosx + sinx } } \, dx = \int\limits {\frac{1}{t} } \, dt$

                      = $ln(t) + C$

                      = $ln(cosx + sinx) + C$

$\int\limits {\frac{1-tanx}{1+tanx} } \, dx = ln(cosx + sinx) + C$