题目
Problem
Find the exact value of ∫ 0 π x 2 cos 1 3 x d x \displaystyle\int_0^\pi x^2\cos\frac13x\,\mathrm{d}x ∫ 0 π x 2 cos 3 1 x d x .
(6)
题目中文翻译
求 ∫ 0 π x 2 cos 1 3 x d x \displaystyle\int_0^\pi x^2\cos\frac13x\,\mathrm{d}x ∫ 0 π x 2 cos 3 1 x d x 的精确值。
(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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两次分部积分依次降低多项式次数。三角函数内部的三分之一会使积分系数乘三,务必逐步保留。
答题过程
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Let
I = ∫ 0 π x 2 cos ( x / 3 ) d x . I=\int_0^\pi x^2\cos(x/3)\,\mathrm{d}x. I = ∫ 0 π x 2 cos ( x /3 ) d x .
For the first integration by parts, take u = x 2 u=x^2 u = x 2 and d v = cos ( x / 3 ) d x \mathrm{d}v=\cos(x/3)\,\mathrm{d}x d v = cos ( x /3 ) d x . Then d u = 2 x d x \mathrm{d}u=2x\,\mathrm{d}x d u = 2 x d x and v = 3 sin ( x / 3 ) v=3\sin(x/3) v = 3 sin ( x /3 ) , giving
I = [ 3 x 2 sin ( x / 3 ) ] 0 π − 6 ∫ 0 π x sin ( x / 3 ) d x . \begin{align*}
I=&\,\big[3x^2\sin(x/3)\big]_0^\pi\\[4mm]
&\,\hspace{2pt}-6\int_0^\pi x\sin(x/3)\,\mathrm{d}x.
\end{align*} I = [ 3 x 2 sin ( x /3 ) ] 0 π − 6 ∫ 0 π x sin ( x /3 ) d x .
For the remaining integral, take u = x u=x u = x and d v = sin ( x / 3 ) d x \mathrm{d}v=\sin(x/3)\,\mathrm{d}x d v = sin ( x /3 ) d x , so v = − 3 cos ( x / 3 ) v=-3\cos(x/3) v = − 3 cos ( x /3 ) . Hence
∫ 0 π x sin ( x / 3 ) d x = [ − 3 x cos ( x / 3 ) ] 0 π + 3 ∫ 0 π cos ( x / 3 ) d x = [ − 3 x cos ( x / 3 ) + 9 sin ( x / 3 ) ] 0 π . \begin{align*}
\int_0^\pi x\sin(x/3)\,\mathrm{d}x
=&\,\big[-3x\cos(x/3)\big]_0^\pi\\[4mm]
&\,\hspace{2pt}+3\int_0^\pi\cos(x/3)\,\mathrm{d}x\\[4mm]
=&\,\big[-3x\cos(x/3)+9\sin(x/3)\big]_0^\pi.
\end{align*} ∫ 0 π x sin ( x /3 ) d x = = [ − 3 x cos ( x /3 ) ] 0 π + 3 ∫ 0 π cos ( x /3 ) d x [ − 3 x cos ( x /3 ) + 9 sin ( x /3 ) ] 0 π .
Substituting this result into I I I ,
I = [ 3 x 2 sin ( x / 3 ) + 18 x cos ( x / 3 ) − 54 sin ( x / 3 ) ] 0 π . \begin{align*}
I=&\,\big[3x^2\sin(x/3)+18x\cos(x/3)\\[4mm]
&\,\hspace{2pt}-54\sin(x/3)\big]_0^\pi.
\end{align*} I = [ 3 x 2 sin ( x /3 ) + 18 x cos ( x /3 ) − 54 sin ( x /3 ) ] 0 π .
At x = 0 x=0 x = 0 , all terms vanish. At x = π x=\pi x = π , use sin ( π / 3 ) = 3 / 2 \sin(\pi/3)=\sqrt3/2 sin ( π /3 ) = 3 /2 and cos ( π / 3 ) = 1 / 2 \cos(\pi/3)=1/2 cos ( π /3 ) = 1/2 :
I = 3 3 2 π 2 + 9 π − 27 3 . \boxed{I=\frac{3\sqrt3}{2}\pi^2+9\pi-27\sqrt3}. I = 2 3 3 π 2 + 9 π − 27 3 .