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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?


marked as duplicate by Martin Sleziak, Amzoti, rschwieb, Asaf Karagila, Mark Bennet May 30 '13 at 17:21

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  • $\begingroup$ This is basically the same as this question. (To be more precise, if you know one of the two integrals, it is easy to calculate the other one and vice versa.) $\endgroup$ – Martin Sleziak May 30 '13 at 16:42

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

  • 2
    $\begingroup$ This is very elegant in that it requires no substitutions. $\endgroup$ – user54147 Feb 9 '13 at 12:20
  • $\begingroup$ @user54147 You are using a substitution, $u = \sin(x) + \cos(x)$ in the end $\endgroup$ – Amad27 Aug 11 '15 at 19:58

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 $\cos x$.


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


$\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 $


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