题目
Problem
Find the exact value of ∫ 0 1 6 π x 2 sin 2 x d x \displaystyle\int_0^{\frac16\pi}x^2\sin2x\,\mathrm{d}x ∫ 0 6 1 π x 2 sin 2 x d x
(6)
题目中文翻译
求 ∫ 0 1 6 π x 2 sin 2 x d x \displaystyle\int_0^{\frac16\pi}x^2\sin2x\,\mathrm{d}x ∫ 0 6 1 π x 2 sin 2 x d x
(6)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
解法一
思路
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多项式乘三角函数用两次分部积分，每次让多项式次数降低。代入零下限时余弦项仍有贡献。
答题过程
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First integrate by parts with u = x 2 u=x^2 u = x 2 d v = sin 2 x d x \mathrm{d}v=\sin2x\,\mathrm{d}x d v = sin 2 x d x
I = ∫ 0 π / 6 x 2 sin 2 x d x = [ − x 2 2 cos 2 x ] 0 π / 6 + ∫ 0 π / 6 x cos 2 x d x . \begin{align*}
I=&\,\int_0^{\pi/6}x^2\sin2x\,\mathrm{d}x\\[4mm]
=&\,\bigg[-\frac{x^2}{2}\cos2x\bigg]_0^{\pi/6}\\[4mm]
&\,\hspace{2pt}+\int_0^{\pi/6}x\cos2x\,\mathrm{d}x.
\end{align*} I = = ∫ 0 π /6 x 2 sin 2 x d x [ − 2 x 2 cos 2 x ] 0 π /6 + ∫ 0 π /6 x cos 2 x d x . A second integration by parts, with u = x u=x u = x d v = cos 2 x d x \mathrm{d}v=\cos2x\,\mathrm{d}x d v = cos 2 x d x
∫ 0 π / 6 x cos 2 x d x = [ x 2 sin 2 x ] 0 π / 6 − 1 2 ∫ 0 π / 6 sin 2 x d x = [ x 2 sin 2 x + 1 4 cos 2 x ] 0 π / 6 . \begin{align*}
\int_0^{\pi/6}x\cos2x\,\mathrm{d}x
=&\,\bigg[\frac{x}{2}\sin2x\bigg]_0^{\pi/6}\\[4mm]
&\,\hspace{2pt}-\frac12\int_0^{\pi/6}\sin2x\,\mathrm{d}x\\[4mm]
=&\,\bigg[\frac{x}{2}\sin2x+\frac14\cos2x\bigg]_0^{\pi/6}.
\end{align*} ∫ 0 π /6 x cos 2 x d x = = [ 2 x sin 2 x ] 0 π /6 − 2 1 ∫ 0 π /6 sin 2 x d x [ 2 x sin 2 x + 4 1 cos 2 x ] 0 π /6 . Therefore
I = [ − x 2 2 cos 2 x + x 2 sin 2 x + 1 4 cos 2 x ] 0 π / 6 . I=\bigg[-\frac{x^2}{2}\cos2x
+\frac{x}{2}\sin2x+\frac14\cos2x\bigg]_0^{\pi/6}. I = [ − 2 x 2 cos 2 x + 2 x sin 2 x + 4 1 cos 2 x ] 0 π /6 . At x = π / 6 x=\pi/6 x = π /6
− π 2 144 + 3 π 24 + 1 8 . -\frac{\pi^2}{144}+\frac{\sqrt3\pi}{24}+\frac18. − 144 π 2 + 24 3 π + 8 1 . At x = 0 x=0 x = 0 1 / 4 1/4 1/4
I = − π 2 144 + 3 π 24 − 1 8 . \boxed{I=-\frac{\pi^2}{144}+\frac{\sqrt3\pi}{24}-\frac18}. I = − 144 π 2 + 24 3 π − 8 1 . 