Question

\( \int_{0}^{\pi / 4} x^{2} \sin 2 x d x= \)

• A. \( \frac{\pi^{2}-2}{8} \)
• B. \( \frac{\pi(\pi-2)}{8} \)
• C. \( \frac{\pi-2}{8} \)
• D. \( \frac{\pi+2}{8} \)

Hint:

Tabular integration reduces calculation errors and writing effort for integrals of the form \( x^n \sin ax \) or \( x^n e^{ax} \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

We evaluate this integral using Integration by Parts (twice) or the Tabular Method (DI Method).
Step 2: Key Formula or Approach:

Tabular Method for \( \int u v dx \): Differentiate \( u \) to 0, Integrate \( v \), multiply diagonals with alternating signs (+, -, +).
Step 3: Detailed Explanation:

Let \( u = x^2 \) and \( dv = \sin 2x dx \). Table: Sign | Differentiate (\(x^2\)) | Integrate (\(\sin 2x\)) + | \( x^2 \) | \( \sin 2x \) - | \( 2x \) | \( -\frac{1}{2}\cos 2x \) + | \( 2 \) | \( -\frac{1}{4}\sin 2x \) - | \( 0 \) | \( \frac{1}{8}\cos 2x \) Result: \( I = \left[ -x^2 \frac{\cos 2x}{2} + 2x \frac{\sin 2x}{4} + 2 \frac{\cos 2x}{8} \right]_0^{\pi/4} \) \( I = \left[ -\frac{x^2}{2}\cos 2x + \frac{x}{2}\sin 2x + \frac{1}{4}\cos 2x \right]_0^{\pi/4} \) Upper limit (\( \pi/4 \)): \( \cos(2 \cdot \pi/4) = \cos(\pi/2) = 0 \). \( \sin(2 \cdot \pi/4) = \sin(\pi/2) = 1 \). Value: \( 0 + \frac{\pi/4}{2}(1) + 0 = \frac{\pi}{8} \). Lower limit (\( 0 \)): \( \cos(0) = 1 \), \( \sin(0) = 0 \). Value: \( 0 + 0 + \frac{1}{4}(1) = \frac{1}{4} \). Total Integral \( I = \frac{\pi}{8} - \frac{1}{4} = \frac{\pi - 2}{8} \).
Step 4: Final Answer:

Matches Option (C).