<s>[INST] Solve the differential equation $p=\sin(10x+6y)$ 
$$p=\sin(10x+6y),p:=dy/dx$$


What I did:
$z=10x+6y\implies 6y=z-10x$
So:
$$dz/dx-10=6\sin z\\
dz/dx=2(5+3\sin z)\\
\int dz/(5+3\sin z)=2\int dx\\
\int 2dt/(5+5t^2+6t)=\int2dx,t:=\tan z/2\\
\int(1/5)dt/((t+3/5)^2-4/5)=2x+a\\
(1/5)(1/(2(\sqrt{4/5})))\ln((t+3/5-2/\sqrt5)/(t+3/5+2/\sqrt5))=2x+b\\
\frac{1}{4\sqrt5}\ln\left(\frac{10x+6y+3/5-2/\sqrt5}{10x+6y+3/5+2/\sqrt5}\right)=2x+c$$

Given answer:
$$y=\frac13\arctan\left(\frac{5\tan 4x}{4-3\tan 4x}\right)-\frac{5x}3$$
 [/INST] I believe you made a slight mistake in your integration:
$$\int \frac{2}{5t^2+6t+5}\, dt = \frac{2}{5}\int \frac{dt}{\left(t+\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2},$$
from which you get
$$2x = \frac{1}{2}\arctan\left(\frac{5t+3}{4}\right)+C.$$
Continuing from here, you should arrive at the answer.
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