Question

integration 1/1-cosx-sinx dx​

Answer 1

Answer:

Let,I=1/(1-cos x-sin x)

=> We know ,sin2x=2tan x/(1+tan^2x)

cos2x=(1-tan^2x)/(1+tan^2x)

Similarly,sin x=2tan x/2/(1+tan^2 x/2)

& cos x=(1-tan^2 x/2)/(1+tan^2 x/2)

These are called half angle formula of sin and cos.

Using these we can solve the given intregal.

I=1/(1-sin x-cos x) dx

I=1/(1-2tan x/2/(1+tan^2 x/2)-(1-tan^2 x/2)/(1+tan^2 x/2) dx

I=(1+tan^2 x/2)/(2tan^2 x/2-2tan x/2) dx

We know ,(1+tan^2 x/2)=sec^2 x/2

I=(sec^2 x/2)/(2tan^2 x/2-2tan x/2) dx

I=1/2(sec^2 x/2)/(tan^2 x/2-tan x/2) dx

Let, t=tan x/2

dt =(sec^2 x/2)*1/2 dx

I=1/t(t-1)

By partial fraction,1/t(t-1)=-1/t+1/(t-1)

I=-dt/t +dt/(t-1)

I=-log|t|+log|t-1|

I=log|(t-1)/t|

I=log|1-1/t|

Putting the value of t in above equation

I=log|1-1/(tan x/2)|

I=log|1-cot x/2|

This is the answer.