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 $I$

  • 9
    $\begingroup$ 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

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


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 $I$ ?


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.