Question

Evaluate: \[ \int_0^{\frac{\pi}{2}} \sqrt{1 - \cos{4x}} \, dx \]

Hint:

To simplify integrals involving trigonometric functions, use known trigonometric identities and properties of integrals.

Answer 1

To solve this integral, we use the identity for \( 1 - \cos{4x} \): \[ 1 - \cos{4x} = 2\sin^2{2x} \] Thus, the integral becomes: \[ \int_0^{\frac{\pi}{2}} \sqrt{2\sin^2{2x}} \, dx = \sqrt{2} \int_0^{\frac{\pi}{2}} |\sin{2x}| \, dx \] Since \( \sin{2x} \) is positive in the interval \( [0, \frac{\pi}{2}] \), we can remove the absolute value: \[ \sqrt{2} \int_0^{\frac{\pi}{2}} \sin{2x} \, dx \] Now, integrate: \[ \int \sin{2x} \, dx = -\frac{1}{2} \cos{2x} \] Thus, the integral becomes: \[ \sqrt{2} \left[ -\frac{1}{2} \cos{2x} \right]_0^{\frac{\pi}{2}} = \sqrt{2} \left( -\frac{1}{2} \left( \cos{\pi} - \cos{0} \right) \right) \] Substitute the values of \( \cos{\pi} \) and \( \cos{0} \): \[ \sqrt{2} \left( -\frac{1}{2} \left( -1 - 1 \right) \right) = \sqrt{2} \left( \frac{1}{2} \times 2 \right) = \sqrt{2} \] Thus, the value of the integral is: \[ \boxed{\sqrt{2}} \]