Compute $\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx$ 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.
 A: Hint: $\sqrt{2}\sin(x+\pi/4)=\sin x +\cos x$, then substitute $x+\pi/4=z$
A: You can do this without thinking: use the Weierstrass substitution to reduce the integral to a rational function, and integrate that as usual.
A: 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 
$$\begin{eqnarray*}
I &=&-\frac{1}{2}\int \frac{1}{1+u}-\frac{u-1}{u^{2}+1}\,du \\
&=&-\frac{1}{2}\ln \left\vert 1+u\right\vert +\frac{1}{4}\ln \left(
u^{2}+1\right) -\frac{1}{2}\arctan u +C\\
&=&-\frac{1}{2}\ln \left\vert 1+\cot x\right\vert +\frac{1}{4}\ln \left(
\cot ^{2}x+1\right) -\frac{1}{2}\arctan \left( \cot x\right) +C \\
&=&-\frac{1}{2}\ln \left\vert 1+\cot x\right\vert +\frac{1}{4}\ln \left(
\csc ^{2}x\right) +\frac{1}{2}x+\text{ Constant} \\
&=&\frac{1}{2}x-\frac{1}{2}\ln \left\vert \sin x+\cos x\right\vert +\text{
Constant.}
\end{eqnarray*}$$
A: 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. 
A: 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
A: 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 $\sin(x) + \cos(x) = k\cos(\phi+x)$:
$$\int \frac{1}{k}\frac{\sin(x)}{\cos(\phi+x)}\mathrm dx.$$
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.
A: 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 $\frac{A\sin x+B\cos x}{C\sin x+D\cos x}$ where $A$ and $B$ couldn't be zero at the same time (one of them can be zero as in this post), and $C$ and $D$ certainly not both zero at the same time.
A: $$= \frac{1}{2} \cdot \int \frac{\sin(x) + \sin(x)}{\sin(x) + \cos(x)} dx = \frac{1}{2} \int \frac{\sin(x) + \cos(x) + \sin(x) - \cos(x) }{\sin(x) + \cos(x)} dx $$ $$= \frac{x}{2} - \frac{1}{2} \int \frac{\cos(x) - \sin(x)}{\sin(x) + \cos(x)} dx$$ Let $u = \sin(x) + \cos(x)$ $$ \implies \frac{x}{2} - \frac{1}{2} \cdot \log \left| \sin(x) + \cos(x) \right| + C$$
