Question

Solve $(1+\cos x)dy=(1-\cos x)dx$.

• A. $y=\cot \frac{x}{2}-x+C$
• B. $y=\tan \frac{x}{2}-x+C$
• C. $y=\sin \frac{x}{2}-x+C$
• D. None of these

Answer 1

Answer: (B)

$y=\tan \frac{x}{2}-x+C$

$dy=\frac{1-\cos x}{1+\cos x}dx$

Integrating both sides,we get

$y=\int -\frac{\cos x+1-2}{\cos x+1}dx$

$y=\int -dx+\int 2\frac{1}{\cos x+1}dx$

$y=-x+\int \frac{{\sec }^2\frac{x}{2}}{2}dx$

$y=\tan \frac{x}{2}-x+C$