Evaluate $\int\frac{\cos x}{\sin x + \cos x}\,\text{d}x$. [duplicate]

Question

I'm trying to figure out how to take this indefinite integral:

$$ \int\frac{\cos x}{\sin x + \cos x}\,\text{d}x.$$

I tried simplifying and rearranging it, and this is the best I got: $$\int\frac{1}{\tan x + 1 }\,\text{d}x.$$

But I still can't figure out how to integrate from there. I know that it's integrable, as Wolfram Alpha indicates that the integral is $ \frac{1}{2}\big(x+\ln{(\sin x + \cos x)}\big)+C$, but I can't figure out the steps to deriving it. Does anyone know how to evaluate this integral?

Answer 1

Let $I=\int\frac{\cos x}{\sin x+\cos x}\ dx$ and $J=\int \frac{\sin x}{\sin x+\cos x}\ dx$.

Then $I+J=\int\frac{\cos x+\sin x}{\sin x+\cos x}\ dx=x+C_1$ and $I-J=\int\frac{\cos x-\sin x}{\sin x+\cos x}\ dx=\ln |\sin x+\cos x|+C_2$.

Hence $I=\frac{1}{2}((I+J)+(I-J))=\frac{1}{2}(x+\ln |\sin x+\cos x|)+C$.

Answer 2

A general (but not necessarily efficient) tool is the Weierstrass substitution $t=\tan(x/2)$.

Then $\cos x=\frac{1-t^2}{1+t^2}$, $\sin x=\frac{2t}{1+t^2}$, and $dx=\frac{2}{1+t^2}$. Do the substitution. We end up needing the following ugly integral: $$\int \frac{2(1-t^2)}{(1+2t-t^2)(1+t^2)}\,dt.$$ We are integraing a rational function, and it can be done using partial fractions. But it takes some work.

The same idea works in principle for any rational function of $\sin x$ and $\cos x$.

Answer 3

Notice that $$\sin x+\cos x=\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x\right)=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$$ Therefore $$I=\frac{1}{\sqrt{2}}\int\frac{\cos x}{\sin\left(x+\frac{\pi}{4}\right)}dx$$ Now let $x+\frac{\pi}{4}=t$ $$I=\frac{1}{\sqrt{2}}\int\frac{\cos \left(t-\frac{\pi}{4}\right)}{\sin t}dt=\frac{1}{2}\int\frac{\sin t + \cos t}{\sin t}dt=\frac{1}{2}\left(t+\ln {\left|\sin t\right|}\right)+C_1\\=\frac{1}{2}\left(x+\frac{\pi}{4}+\ln {\left|\sin x+\cos x\right|}\right)+C_1=\frac{1}{2}\left(x+\ln {\left|\sin x+\cos x\right|}\right)+C$$

Answer 4

$\displaystyle \int \frac{1}{1+ \tan x} \ dx $

$ \displaystyle = \int \frac{1}{1+u} \frac{1}{1+u^{2}} \ du$ (let $u = \tan x$)

$ \displaystyle = \frac{1}{2} \int \left( \frac{1}{1+u} + \frac{1-u}{1+u^{2}} \right) \ du$

$ \displaystyle = \frac{1}{2} \int \left( \frac{1}{1+u} - \frac{u}{1+u^{2}} + \frac{1}{1+u^{2}} \right) \ du $

$ \displaystyle =\frac{1}{2} \left(\ln(1+u) - \frac{1}{2} \ln(1+u^{2}) + \arctan u \right) + C$

$ \displaystyle =\frac{1}{2} \left(\ln(1+u) - \ln(\sqrt{1+u^{2}}) + \arctan u \right) + C$

$ \displaystyle = \frac{1}{2} \left( \ln(1+\tan x) - \ln (\sqrt{1+\tan^{2}x}) + x \right) + C$

$ \displaystyle = \frac{1}{2} \left( \ln(1+\tan x) - \ln (\sec x) + x \right) + C$

$ \displaystyle = \frac{1}{2} \Big( \ln \big(\frac{1+ \tan x}{\sec x}\big) + x \Big) + C$

$ \displaystyle = \frac{1}{2} \left( \ln (\cos x + \sin x ) + x \right) + C $