I'm having trouble computing the integral: $$\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx.$$ I hope that it can be expressed in terms of elementary functions. I've tried simple substitutions such as $u=\sin(x)$ and $u=\cos(x)$, but it was not very effective.

Any suggestions are welcome. Thanks.


Let $I:=\int\frac{\cos x}{\cos x+\sin x}dx$ and $J:=\int\frac{\sin x}{\cos x+\sin x}dx$. Then $I+J=x + C$, and $$I-J=\int\frac{\cos x-\sin x}{\cos x+\sin x}dx=\int\frac{u'(x)}{u(x)}dx,$$ where $u(x)=\cos x+\sin x$. Now we can conclude.

  • 14
    $\begingroup$ This is almost too cute for its own good :-) $\endgroup$ – Mariano Suárez-Álvarez Aug 9 '12 at 17:06
  • 3
    $\begingroup$ Nice proof. Simple and elegant. Thanks! $\endgroup$ – Michael Li Aug 9 '12 at 17:12
  • 10
    $\begingroup$ You can make this even simpler by just writing the $\sin x$ in the numerator as ${1 \over 2}(\sin x + \cos x) - {1 \over 2}(\cos x - \sin x)$. $\endgroup$ – Zarrax Aug 9 '12 at 17:12
  • 6
    $\begingroup$ @Zarrax, that magical step only helps in hiding what is going on :-) $\endgroup$ – Mariano Suárez-Álvarez Aug 9 '12 at 18:09
  • 17
    $\begingroup$ In this math education article the author describes giving the same problem to a young Terence Tao, aged 8; he gave essentially the same beautiful solution. $\endgroup$ – Erick Wong Aug 9 '12 at 18:29

Hint: $\sqrt{2}\sin(x+\pi/4)=\sin x +\cos x$, then substitute $x+\pi/4=z$

  • 4
    $\begingroup$ Thanks. This is a new formula for me. The pity is a number of constants appears meanwhile. $\endgroup$ – Michael Li Aug 9 '12 at 17:30
  • 3
    $\begingroup$ Very nice, at least if we can regard $\int \csc t \,dt$ as a "book" integral. $\endgroup$ – André Nicolas Aug 9 '12 at 18:57
  • 20
    $\begingroup$ @Michael This formula generalizes to $$A\cos x+B\sin x=\sqrt{A^{2}+B^{2}}\sin \left( x+\arctan \frac{A}{B}\right) ,$$ which is very useful in (electrical) engineering. $\endgroup$ – Américo Tavares Aug 9 '12 at 20:05
  • $\begingroup$ @AméricoTavares Why this is so important in EE I thought this is pretty basic formula. $\endgroup$ – A---B Apr 23 '17 at 15:19
  • $\begingroup$ @A---B Yes, it's pretty basic, but it allows us to use the complex plane to represent currents, voltages and impedances of a.c. circuits operating in a single frequence. V = Z I, where V is a voltage drop across the impedance Z and I is the current through it. $\endgroup$ – Américo Tavares Apr 23 '17 at 16:37

You can do this without thinking: use the Weierstrass substitution to reduce the integral to a rational function, and integrate that as usual.


We can write the integrand as $$\begin{equation*} \frac{1}{1+\cot x} \end{equation*}$$ and use the substitution $u=\cot x$. Since $du=-\left( 1+u^{2}\right) dx$ we reduce it to a rational function

$$\begin{equation*} I:=\int \frac{\sin x}{\sin x+\cos x}dx=-\int \frac{1}{\left( 1+u\right) \left( u^{2}+1\right) }\,du. \end{equation*}$$

By expanding into partial fractions and using the identities

$$\begin{eqnarray*} \cot ^{2}x+1 &=&\csc ^{2}x \\ \arctan \left( \cot x\right) &=&\frac{\pi }{2}-x \\ \frac{\csc x}{1+\cot x} &=&\frac{1}{\sin x+\cos x} \end{eqnarray*}$$

we get

$$\begin{eqnarray*} I &=&-\frac{1}{2}\int \frac{1}{1+u}-\frac{u-1}{u^{2}+1}\,du \\ &=&-\frac{1}{2}\ln \left\vert 1+u\right\vert +\frac{1}{4}\ln \left( u^{2}+1\right) -\frac{1}{2}\arctan u +C\\ &=&-\frac{1}{2}\ln \left\vert 1+\cot x\right\vert +\frac{1}{4}\ln \left( \cot ^{2}x+1\right) -\frac{1}{2}\arctan \left( \cot x\right) +C \\ &=&-\frac{1}{2}\ln \left\vert 1+\cot x\right\vert +\frac{1}{4}\ln \left( \csc ^{2}x\right) +\frac{1}{2}x+\text{ Constant} \\ &=&\frac{1}{2}x-\frac{1}{2}\ln \left\vert \sin x+\cos x\right\vert +\text{ Constant.} \end{eqnarray*}$$


Write the numerator (here $\sin x$) as a linear combination of the denominator and the derivative of the denominator: $$A(\sin x+ \cos x) + B( \cos x- \sin x) = \sin x$$ Solve for $A$ and $B$ and split the fraction accordingly. Integrating give a linear term and an $\ln$ This method generally works for $\frac{Asinx+Bcosx}{Csinx+Dcosx}$ where $A$ and $B$ not both zero (one of them can be zero as in this post), and $C$ and $D$ certainly not both zero at the same time.


$$= \frac{1}{2} \cdot \int \frac{\sin(x) + \sin(x)}{\sin(x) + \cos(x)} dx = \frac{1}{2} \int \frac{\sin(x) + \cos(x) + \sin(x) - \cos(x) }{\sin(x) + \cos(x)} dx $$ $$= \frac{x}{2} - \frac{1}{2} \int \frac{\cos(x) - \sin(x)}{\sin(x) + \cos(x)} dx$$ Let $u = \sin(x) + \cos(x)$ $$ \implies \frac{x}{2} - \frac{1}{2} \cdot \log \left| \sin(x) + \cos(x) \right| + C$$

  • $\begingroup$ related: math.stackexchange.com/questions/1362988/… $\endgroup$ – Math-fun Aug 11 '15 at 20:03
  • $\begingroup$ Great, but this answer was already given, 3 years ago, in the comment to the Top Answer. $\endgroup$ – Mark Viola Aug 11 '15 at 20:06
  • $\begingroup$ @Dr.MV, ah darn! I missed that somehow, sorry $\endgroup$ – Amad27 Aug 11 '15 at 20:21
  • 1
    $\begingroup$ @amad27 No worry. Happens often on MSE. It's happened to me before. Be well my friend. ;-) And +1 anyway $\endgroup$ – Mark Viola Aug 11 '15 at 20:26

Just for some new ideas! I would reccomend a completely different method. This method uses the Gudermannian $\text{gd}$ function. So you would substitute $x=\text{gd}(a);\text{d}x=\text{sech}\space a\text{d}a$ That transforms the integral into:

$$\int \frac{\tanh a}{\tanh a+\text{sech}\space a}(\text{sech}\space a)\mathrm da$$

Through some hyperbolic trig properties, we get (correct me if I'm wrong)

$$\int {\frac{1}{\cosh a+\coth a}}\text{d}a$$

You could probably take it from here

  • $\begingroup$ There are plenty of awesome answers out here, just wanted to offer a new way $\endgroup$ – user285523 Nov 29 '15 at 22:13
  • $\begingroup$ I am giving you a bounty because the idea is really impressive:i.e. new $\endgroup$ – user266519 Dec 1 '15 at 18:33

So we have the integral

$$\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx.$$

A little bit more general solution would be to substitute $\sin(x) + \cos(x) = \cos(\phi+x)$:

$$\int \frac{\sin(x)}{\cos(\phi+x)}\mathrm dx.$$

We know this is true from elementary trigonometry. Now if we know the logarithmic derivative : $$\frac{\partial \ln(g(x))} {\partial x} = \frac{g'(x)}{g(x)}$$ we are almost done immediately without even needing to do any substitution, however of course we need to determine $\phi$ and add a constant of integration, but that is a rather easy exercise.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.