Integral of $\int\frac{\sin x}{\sin x+\cos x}dx$

Question

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 $I$

If you know what $J-I$ is, you know what $I-J$ is. Try adding $I-J$ and $I+J$ together. — welshman500, Dec 12 '18 at 17:36
Another post about the same integral: Compute $\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx$. — Martin Sleziak, Feb 18 '21 at 08:46

score 2 · Answer 1 · answered Dec 12 '18 at 17:45

2

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 $I$ ?

answered Dec 12 '18 at 17:45

Rebellos

21,666

score 0 · Accepted Answer · answered Dec 12 '18 at 19:44

0

what you have is: $$I+J=x+C$$ $$I-J=-\ln|\cos(x)+\sin(x)|+C$$ so: $$2I=x-\ln|\cos(x)+\sin(x)|+C$$

answered Dec 12 '18 at 19:44

Henry Lee

12,554

Integral of $\int\frac{\sin x}{\sin x+\cos x}dx$

2 Answers2

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