Integration?

Question

integrate cos x / (cos x + sin x) dx and sinx /(cos x + sin x) dx

I know it should be simple but, i have not integrated for a while

I have multiplied by through but, how do i get to 1/1+ tan^2

Answer 1

we have cos x / (cos x + sin x) = 1/(1 + tanx). To integrate this, put x = arctan t, so that dx = dt/(1 + t^2). We get

Int 1/(1 + tanx) dx =. We know from Algebra that 1/((1 + t)(1+ t^2)) = a/(1 +t) + (bt + c)/(1 + t^2). Then, a(1 + t^2) + (1 +t)(bt + c) = 1 => a + at^2 + bt + c + bt^2 + ct = 1 => (a + b) t^2 + (b + c) t + a + c =1. Then,

a + b = 0 => a = -b
b + c = 0 => c = -b
a+ c =1

So, -b - b =1 and b =-1/2, . a = c =1/2 .

Now, we can split into 3 integrals

Int 1/((1 + t)(1+ t^2)) dt = (1/2) Int dt(1+t) - 1/2 Int t/(1 + t^2) dt + (1/2) Intdt/(1 + t^2) = 1/2 ln (1 + t) -1/4 ln(1 + t^2) +1/2 arctant + C.

Since x = arctan t, t = tan x and we finaly have Int cos x / (cos x + sin x) dx = 1/2 ln(1 + tan x) -1/4 ln(1 + tan^2 x) +1/2 x + C = 1/2 ln(1 + tan x) -1/2 ln(sec x) +1/2 x + C , since 1 + tan^2 x =sec^2 x

The other integral is done similarly.

Answer 2

In principle f(x) = [1/f(x)]^-1

Thus cos x / (cos x + sin x) = [(cos x + sin x)/cos x]^-1
= 1/[1+tan x]

Integral of 1/[1+tan x] would be ln[1+tan x]/(sec x)^2 + c

Answer 3

try multiplying by the complex conjugate.

(cos x - sin x) on top and bottom. then turn cos^2 into sin^2 or vice versa, whichever is easier.