Question

The value of $\int_0^{\pi/2} \sin^2 \frac{x}{2} \cos^2 \frac{x}{2} \, dx$ is given by}

• A. $\pi/2$
• B. $\pi/4$
• C. $\pi/16$
• D. $\pi$

Hint:

The average value of $\sin^2 x$ over a complete period or half-period (like $[0, \pi/2]$) is $\frac{1}{2}$. Thus, $\int_0^{\pi/2} \frac{1}{4} \sin^2 x \, dx = \frac{1}{4} \times \left( \frac{1}{2} \times \frac{\pi}{2} \right) = \frac{\pi}{16}$ directly.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem requires simplifying the integrand using double-angle trigonometric identities before performing the definite integration.
Key Formula or Approach:
Use the identity for the sine of a double angle: \[ \sin \theta \cos \theta = \frac{1}{2} \sin 2\theta \] And the power-reducing identity: \[ \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \]

Step 2: Detailed Explanation:
Let the given integral be $I$: \[ I = \int_0^{\pi/2} \sin^2 \frac{x}{2} \cos^2 \frac{x}{2} \, dx \]
We rewrite the integrand as: \[ \sin^2 \frac{x}{2} \cos^2 \frac{x}{2} = \left( \sin \frac{x}{2} \cos \frac{x}{2} \right)^2 = \left( \frac{1}{2} \sin x \right)^2 = \frac{1}{4} \sin^2 x \]
Substitute this back into the integral: \[ I = \int_0^{\pi/2} \frac{1}{4} \sin^2 x \, dx = \frac{1}{4} \int_0^{\pi/2} \left( \frac{1 - \cos 2x}{2} \right) \, dx \] \[ = \frac{1}{8} \int_0^{\pi/2} (1 - \cos 2x) \, dx \] \[ = \frac{1}{8} \left[ x - \frac{\sin 2x}{2} \right]_0^{\pi/2} \]
Evaluate this expression at the upper and lower limits: \[ I = \frac{1}{8} \left[ \left( \frac{\pi}{2} - \frac{\sin \pi}{2} \right) - \left( 0 - \frac{\sin 0}{2} \right) \right] \] \[ = \frac{1}{8} \left[ \frac{\pi}{2} - 0 \right] = \frac{\pi}{16} \]

Step 3: Final Answer:
The value of the definite integral is $\pi/16$.