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The questions defines $$I=\int\frac{\sin x}{\sin x +\cos x}dx\;\;J=\int\frac{\cos x}{\sin x +\cos x}dx$$ It asked me to find $I+J$ and $J-I$ which I have done and I will show below but now I need to find the integral shown below and I'm unsure on what to do. $$\int\frac{\sin x}{\sin x+\cos x}dx$$

I have found that: $$I+J = x+c$$ $$J-I=\ln{|\cos x +\sin x|} +c$$

But now i'm unsure on how to find just $I$

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    $\begingroup$ If you know what $J-I$ is, you know what $I-J$ is. Try adding $I-J$ and $I+J$ together. $\endgroup$ – welshman500 Dec 12 '18 at 17:36
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what you have is: $$I+J=x+C$$ $$I-J=-\ln|\cos(x)+\sin(x)|+C$$ so: $$2I=x-\ln|\cos(x)+\sin(x)|+C$$

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Hint :

Consider the following system of equations :

$$\begin{cases} J+I = x+c \\ J - I = \ln|\cos x + \sin x | + c\end{cases}$$

See an easy way out for $I$ ?

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