Question

Evaluate the integral $ I = \int \frac{\sin 4x}{\sin 2x} \, dx $

• A. \( 2 \)
• B. \( 4 \)
• C. \( 1 \)
• D. \( 0 \)

Hint:

When dealing with integrals involving trigonometric functions, simplify using known identities and make sure to check the integration constants.

Answer 1

The Correct Option is B

We are given the integral: \[ I = \int \frac{\sin 4x}{\sin 2x} \, dx \] This can be simplified using a known trigonometric identity for \( \sin 4x \) in terms of \( \sin 2x \): \[ \sin 4x = 2 \sin 2x \cos 2x \] Substituting this into the integral: \[ I = \int \frac{2 \sin 2x \cos 2x}{\sin 2x} \, dx \] The \( \sin 2x \) terms cancel out, leaving: \[ I = \int 2 \cos 2x \, dx \] Now integrate \( 2 \cos 2x \): \[ I = 2 \times \frac{\sin 2x}{2} = \sin 2x \]
Thus, the answer is \( 4 \).