Solve $\frac{{dy}}{{dx}} = \cos (x + y) + \sin (x + y)$.

Given, $\frac{{dy}}{{dx}} = \cos (x + y) + \sin (x + y)$

…..(i)
Put $x + y = z$

$\Rightarrow$ $1 + \frac{{dy}}{{dx}} = \frac{{dz}}{{dx}}$

On substituting these values in Eq. (i),

we get $\left( {\frac{{dz}}{{dx}} - 1} \right) = \cos z + \sin z$

$\Rightarrow$ $\frac{{dz}}{{dx}} = (\cos z + \sin z + 1)$

$\Rightarrow$ $\frac{{dz}}{{\cos z + \sin z + 1}} = dx$

On integrating both sides,

we get
$\int {\frac{{dz}}{{\cos z + \sin z + 1}}} = \int 1 dx$

$\Rightarrow$ $\int {\frac{{dz}}{{\frac{{1 - {{\tan }^2}z/2}}{{1 + {{\tan }^2}z/2}} + \frac{{2\tan z/2}}{{1 + {{\tan }^2}z/2}} + 1}}} = \int d x$

$\Rightarrow$ $\int {\frac{{dz}}{{\frac{{1 - {{\tan }^2}z/2 + 2\tan z/2 + 1 + {{\tan }^2}z/2}}{{\left( {1 + {{\tan }^2}z/2} \right)}}}}} = \int d x$

$\Rightarrow$ $\int {\frac{{\left( {1 + {{\tan }^2}z/2} \right)dz}}{{2 + 2{{\tan }^2}z/2}}} = \int d x$

$\Rightarrow$ $\int {\frac{{{{\sec }^2}z/2dz}}{{2(1 + \tan z/2)}}} = \int d x$

Put $1 + \tan z/2 = t \Rightarrow \left( {\frac{1}{2}{{\sec }^2}z/2} \right)dz = dt$

$\Rightarrow$ $\int {\frac{{dt}}{t}} = \int d x$

$\Rightarrow$ $\log |t| = x + C$

$\Rightarrow$ $\log |1 + \tan z/2| = x + C$

$\Rightarrow$ $\log \left| {1 + \tan \frac{{(x + y)}}{2}} \right| = x + C$