Question

Evaluate: \( \int_0^{\pi/2} (1 - \cos 4x) \, dx \).

Hint:

Use trigonometric identities to simplify integrals; evaluate definite integrals by applying limits.

Answer 1

Use identity: \( 1 - \cos 4x = 2 \sin^2 2x \).
\[ \int_0^{\pi/2} (1 - \cos 4x) \, dx = \int_0^{\pi/2} 2 \sin^2 2x \, dx = 2 \int_0^{\pi/2} \sin^2 2x \, dx. \] Use \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \): \[ \sin^2 2x = \frac{1 - \cos 4x}{2}, \quad \text{but directly integrate } \sin^2 2x. \] \[ \int \sin^2 2x \, dx = \int \frac{1 - \cos 4x}{2} \, dx = \frac{1}{2} \left( x - \frac{\sin 4x}{4} \right) + c. \] Evaluate: \[ 2 \cdot \frac{1}{2} \left[ x - \frac{\sin 4x}{4} \right]_0^{\pi/2} = \left[ x - \frac{\sin 4x}{4} \right]_0^{\pi/2} = \left( \frac{\pi}{2} - \frac{\sin 2\pi}{4} \right) - (0 - 0) = \frac{\pi}{2}. \] Answer: \( \frac{\pi}{2} \).