Question

http://www.imagebam.com/image/7c8b87441089411
anyone can tell me how to change it reseparable to separable then solve it?

Answer 1

try a sub \(v = x+y\), so \(v' = 1 + y'\) and just process it from there....

Answer 2

ok i did it
\[\frac{ dy }{ dx}=\sin (x+y)\]
\[(x+y=u )Dervitave \to x (1+\frac{ dy }{ dx }=\frac{ du }{ dx})\]
then \[1+\sin u=\frac{ du }{ dx }\]
\[dx=\frac{ 1 }{ 1+\sin u}du\]
times it's conjucate
dx=\[\frac{ 1 }{ 1+\sin (u) }*\frac{ 1-\sin (u) }{ 1-\sin (u) }\]
but i stopped there can anyone help

Answer 3

that looks good

you now have \(\large \int \frac{1 - \sin u}{\cos^2 u} = \int \sec^2 u - \sec u \tan u\)

you know those...

Answer 4

thank you it's all clear now