Question

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

• A. $\ln [1+\tan \frac{x+y}{2}]=x+c$
• B. $\ln [1+\tan \frac{x+y}{2}]=2x-c$
• C. $\ln \left[\tan \frac{x+y}{2}\right]=x+2c$
• D. $\ln [1-\tan \frac{x+y}{2}]=x-2c$

Answer 1

The correct option is A $\ln [1+\tan \frac{x+y}{2}]=x+c$

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

Put $u=x+y\Rightarrow y=u-x\Rightarrow \frac{dy}{dx}=\frac{du}{dx}-1$

$\therefore \frac{du}{dx}-1=\sin u+\cos u\Rightarrow \frac{du}{dx}=(1+\cos u)+\sin u$

$\Rightarrow \frac{du}{dx}=2{\cos }^2\frac{u}{2}+2\sin \frac{u}{2}\cos \frac{u}{2}$

$\Rightarrow \frac{1}{2}{\sec }^2\frac{u}{2}\frac{du}{dx}=1+\tan \frac{u}{2}$
$\Rightarrow \frac{1}{2}\left(\frac{{\sec }^2\frac{u}{2}}{\tan \frac{u}{2}+1}\right)du=dx$

Integrate on both sides

$\Rightarrow \log (1+\tan \frac{u}{2})=x+c$
$\Rightarrow \log (1+\tan \frac{x+y}{2})=x+c$