Help with an integral

Question

What is the antiderivative of (sin x)/(cos x + sin x)?

I am able to solve this using the substitution u=tan(x/2).

Can anyone give me a simpler method?

Answer 1

f(x) = sinx/(sinx+cosx)

= 1/(1+cosx/sinx) = 1/(1+cotx)

= (1+cotx)^-1

F(x) = what you get by raising the power by 1, dividing by that result and multiplying that by derivative of cotx

but that leads to division by zero, not allowed. So do the natural log alternative

F(x) = (x-ln|sinx + cosx|)/2 + C

use sinx = (sinx+cosx)/2 - (cosx-sinx)/2

then sinx/(sinx+cosx) = 1/2 - (cosx-sinx)/2(sinx+cosx)

let u=sinx +cosx

1/2 integral of (sinx-cosx)/(sinx+cosx) = - integral of du/u =-lnu = -ln(sinx+cosx)

integral of 1/2 - (cosx-sinx)/2(sinx+cosx) = x/2 - [ln|sinx+cosx|]/2 + C

Answer 2

Answer 3

The question has already been answered but the key is to recognize that a substitution of u = ln(sinx+cosx) results in du = cosx-sinx/(cosx+sinx). Therefore you should attempt to manipulate the integrand into a form that matches the expression for du with cosx-sinx in the numerator. As has already been pointed out, this can be done by expressing sinx as (sinx+cosx/)2 - (cosx-sinx)/2.

This results in an integrand of (1 - (cosx-sinx)/(cosx+sinx))/2 and after the aforementioned u substitution, you get an integral of 1/2dx - 1/2du which gives you x/2 - u/2.