TS EAMCET · Maths · Definite Integration
\(\int_0^{\pi / 4} x^2 \sin 2 x d x=\)
- A \(\frac{\pi^2-2}{8}\)
- B \(\frac{\pi(\pi-2)}{8}\)
- C \(\frac{\pi-2}{8}\)
- D \(\frac{\pi+2}{8}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi-2}{8}\)
Step-by-step Solution
Detailed explanation
\( \int x^2 \sin 2x \, dx = -\frac{x^2}{2} \cos 2x + \frac{x}{2} \sin 2x + \frac{1}{4} \cos 2x \) \( \left[-\frac{x^2}{2} \cos 2x + \frac{x}{2} \sin 2x + \frac{1}{4} \cos 2x \right]_0^{\pi/4} \)…
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