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

$\ln|1 + \tan(\frac{x+y}{2})| = x + C$

The correct answer is Option (1) → $\ln|1 + \tan(\frac{x+y}{2})| = x + C$ ##

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

Put $x + y = z ⇒1 + \frac{dy}{dx} = \frac{dz}{dx}$

On substituting these values in Eq. (i), we get

$\left( \frac{dz}{dx} - 1 \right) = \cos z + \sin z \quad \left[ \frac{dy}{dx} = \frac{dz}{dx} - 1 \right]$

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

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

On integrating both sides, we get

$\int \frac{dz}{\cos z + \sin z + 1} = \int 1 dx \quad \text{[using variable separable]}$

$⇒\int \frac{dz}{\frac{1 - \tan^2 z/2}{1 + \tan^2 z/2} + \frac{2 \tan z/2}{1 + \tan^2 z/2} + 1} = \int dx \quad \left[ ∵\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right]$

$⇒\int \frac{dz}{\frac{1 - \tan^2 z/2 + 2 \tan z/2 + 1 + \tan^2 z/2}{(1 + \tan^2 z/2)}} = \int dx \quad \left[ \sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} \right]$

$⇒\int \frac{(1 + \tan^2 z/2) dz}{2 + 2 \tan z/2} = \int dx$

$⇒\int \frac{\sec^2 z/2 \, dz}{2(1 + \tan z/2)} = \int dx$

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

$⇒\int \frac{dt}{t} = \int dx$

$⇒\log |t| = x + C \quad \left[ ∵t = 1 + \tan \frac{z}{2} \text{ and } z = x + y \right]$

$⇒\log |1 + \tan z/2| = x + C$

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