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:
$$y=\frac13\arctan\left(\frac{5\tan 4x}{4-3\tan 4x}\right)-\frac{5x}3$$

The differential equation write $$\frac{dy}{dx}=\sin(10x+6y)$$ Just as you did, defining $z=10x+6y$, we then have after some rearrangements $$\frac{dz}{dx}=6\sin(z)+10$$ which is separable; so$$\frac{dx}{dz}=\frac{1}{6\sin(z)+10}$$ $$x+c=\int\frac{dz}{6\sin(z)+10}$$ Using Weierstrass substitution, we then arrive to $$x+c=\int\frac{dt}{5 t^2+6 t+5}=\frac{1}{5}\int\frac{dt}{t^2+\frac65 t+1}=\frac{1}{5}\int\frac{dt}{(t+\frac35)^2 +\frac{6}{15}}$$ This is the place where I think there is a small mistake.
So, $$x+c=\frac{1}{4} \tan ^{-1}\left(\frac{1}{4} (5 t+3)\right)$$
I am sure that you can take from here and finish the problem.
