Question

The solution of \(\frac{\text{d}y}{\text{d}x} = sin(x+y)+cos(x+y)\) is

• A. \(log[1+tan(\frac{x+y}{2})] = y+c\), where c is the constant of integration
• B. \(log[1-tan(\frac{x+y}{2})] = y+c\), where c is the constant of integration
• C. \(log[1+tan(\frac{x+y}{2})] = x+c\), where c is the constant of integration
• D. \(log[1-tan(\frac{x+y}{2})] = x+c\), where c is the constant of integration

Hint:

Substitute v = x + y and use the tangent half angle substitution.

Answer 1

The Correct Option is C

Step 1: New variable
$v = x + y$ gives $v' = 1 + \sin v + \cos v$.

Step 2: Identity
$1 + \cos v = 2\cos^2\frac v2$ and $\sin v = 2\sin\frac v2\cos\frac v2$, so $1 + \sin v + \cos v = 2\cos\frac v2\left(\cos\frac v2 + \sin\frac v2\right)$.

Step 3: Integrate
$dx = \dfrac{dv}{2\cos\frac v2(\cos\frac v2 + \sin\frac v2)} = \dfrac{\frac12\sec^2\frac v2\,dv}{1 + \tan\frac v2}$, so $x + c = \log\left(1 + \tan\frac v2\right)$. Option (C).

Final Answer:
Option (C). \[ \boxed{\log\left[1 + \tan\left(\frac{x+y}{2}\right)\right] = x + c} \]