_ 0 ^ x^ 2 2 x [( / 2) x] 2 x- d x=
Mathematics · JEE Advanced · NTA Exams — Integral Calculus
\(\int_{0}^{\pi} \frac{x^{2} \sin 2 x \sin [(\pi / 2) \cos x]}{2 x-\pi} d x=\)
- \(\frac{4}{\pi^{2}}\)
- \(\frac{\pi^{2}}{4}\)
- \(\frac{\pi^{2}}{8}\)
Answer
(D)
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