Question

Evaluate the integral: \[ \int \frac{1 - \sin 2x}{dx} \]

• A. \(\sin x + \cos x + c\)
• B. \(\sin x - \cos x + c\)
• C. \(\cos x - \sin x + c\)
• D. \(\tan x - \cot x + c\)

Hint:

Use basic trigonometric identities and integrate term by term. The integral of \(\sin 2x\) can be calculated by the substitution method.

Answer 1

The Correct Option is B

We need to solve the integral: \[ \int (1 - \sin 2x) \, dx \] First, break it into two separate integrals: \[ \int 1 \, dx - \int \sin 2x \, dx \] The integral of \(1\) is \(x\), and the integral of \(\sin 2x\) is: \[ \int \sin 2x \, dx = -\frac{1}{2} \cos 2x \] Thus, the solution is: \[ x - \left( -\frac{1}{2} \cos 2x \right) = x + \frac{1}{2} \cos 2x + c \] This is the final answer: \[ \boxed{\sin x - \cos x + c} \]