Question

Evaluate: $\frac{dy}{dx}=\sin (x+y)+\cos (x+y)$

Answer 1

$\begin{array}{c}\frac{dy}{dx}=\sin \left(x+y\right)+\cos \left(x+y\right)\\ putx+y=t\\ 1+\frac{dy}{dx}=\frac{dt}{dx}\\ \frac{dt}{dx}=\sin t+\cos t+1\\ \int \frac{1}{1+\sin t+\cos t}dt=\int dx+C\\ =\int \frac{{\sec }^2\frac{t}{2}st}{1+{\tan }^2\frac{t}{2}+2\tan \frac{t}{2}+1-{\tan }^2\frac{t}{2}}=x+c\\ =\int \frac{{\sec }^2\frac{t}{2}dt}{1+2\tan \frac{t}{2}}=x+c\\ =\log \left\{1+2\tan \left(\frac{x+y}{2}\right)\right\}=x+c\end{array}$