题目
Problem
Let I = ∫ 1 3 x 3 3 + x 2 d x I=\displaystyle\int_1^3\frac{x^3}{3+x^2}\,\mathrm{d}x I = ∫ 1 3 3 + x 2 x 3 d x
(a) Using the substitution x = 3 tan u x=\sqrt3\tan u x = 3 tan u I = ∫ 1 6 π 1 3 π 3 tan 3 u d u I=\displaystyle\int_{\frac16\pi}^{\frac13\pi}3\tan^3u\,\mathrm{d}u I = ∫ 6 1 π 3 1 π 3 tan 3 u d u
(4)
(b) Hence, or otherwise, find the exact value of I I I p + q ln r p+q\ln r p + q ln r p p p q q q r r r
(5)
题目中文翻译
设 I = ∫ 1 3 x 3 3 + x 2 d x I=\displaystyle\int_1^3\frac{x^3}{3+x^2}\,\mathrm{d}x I = ∫ 1 3 3 + x 2 x 3 d x
(a) 使用代换 x = 3 tan u x=\sqrt3\tan u x = 3 tan u I = ∫ 1 6 π 1 3 π 3 tan 3 u d u I=\displaystyle\int_{\frac16\pi}^{\frac13\pi}3\tan^3u\,\mathrm{d}u I = ∫ 6 1 π 3 1 π 3 tan 3 u d u
(4)
(b) 由此或使用其他方法，求 I I I p + q ln r p+q\ln r p + q ln r p p p q q q r r r
(5)
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
(a)
解法一
思路
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换元时同时转换分子、分母、微分和上下限，最后约掉公共因子。
答题过程
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With x = 3 tan u x=\sqrt3\tan u x = 3 tan u
d x = 3 sec 2 u d u , 3 + x 2 = 3 ( 1 + tan 2 u ) = 3 sec 2 u . \begin{align*}
\mathrm{d}x=&\,\sqrt3\sec^2u\,\mathrm{d}u,\\[4mm]
3+x^2=&\,3(1+\tan^2u)=3\sec^2u.
\end{align*} d x = 3 + x 2 = 3 sec 2 u d u , 3 ( 1 + tan 2 u ) = 3 sec 2 u . The new limits satisfy
x = 1 ⟹ tan u = 1 3 ⟹ u = π 6 , x=1\ \Longrightarrow\ \tan u=\frac1{\sqrt3}
\ \Longrightarrow\ u=\frac\pi6, x = 1 ⟹ tan u = 3 1 ⟹ u = 6 π , and
x = 3 ⟹ tan u = 3 ⟹ u = π 3 . x=3\ \Longrightarrow\ \tan u=\sqrt3
\ \Longrightarrow\ u=\frac\pi3. x = 3 ⟹ tan u = 3 ⟹ u = 3 π . Hence
I = ∫ π / 6 π / 3 3 3 tan 3 u 3 sec 2 u 3 sec 2 u d u = ∫ π / 6 π / 3 3 tan 3 u d u . \begin{align*}
I=&\,\int_{\pi/6}^{\pi/3}
\frac{3\sqrt3\tan^3u}{3\sec^2u}
\sqrt3\sec^2u\,\mathrm{d}u\\
=&\,\boxed{\int_{\pi/6}^{\pi/3}3\tan^3u\,\mathrm{d}u}.
\end{align*} I = = ∫ π /6 π /3 3 sec 2 u 3 3 tan 3 u 3 sec 2 u d u ∫ π /6 π /3 3 tan 3 u d u .
(b)
解法一
思路
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承接(a)，把 tan 3 u \tan^3u tan 3 u tan u ( sec 2 u − 1 ) \tan u(\sec^2u-1) tan u ( sec 2 u − 1 )
答题过程
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From (a), use tan 2 u = sec 2 u − 1 \tan^2u=\sec^2u-1 tan 2 u = sec 2 u − 1 tan u \tan u tan u sec 2 u \sec^2u sec 2 u ∫ tan u d u = − ln ( cos u ) + C \int\tan u\,\mathrm{d}u=-\ln(\cos u)+C ∫ tan u d u = − ln ( cos u ) + C
I = 3 ∫ π / 6 π / 3 tan u ( sec 2 u − 1 ) d u = [ 3 2 tan 2 u + 3 ln ( cos u ) ] π / 6 π / 3 . \begin{align*}
I=&\,3\int_{\pi/6}^{\pi/3}
\tan u(\sec^2u-1)\,\mathrm{d}u\\
=&\,\bigg[\frac32\tan^2u+3\ln(\cos u)\bigg]_{\pi/6}^{\pi/3}.
\end{align*} I = = 3 ∫ π /6 π /3 tan u ( sec 2 u − 1 ) d u [ 2 3 tan 2 u + 3 ln ( cos u ) ] π /6 π /3 . Here cos u > 0 \cos u>0 cos u > 0
I = 3 2 ( 3 − 1 3 ) + 3 ln ( 1 / 2 3 / 2 ) = 4 + 3 ln 1 3 = 4 − 3 2 ln 3 . \begin{align*}
I=&\,\frac32\bigg(3-\frac13\bigg)
+3\ln\bigg(\frac{1/2}{\sqrt3/2}\bigg)\\
=&\,4+3\ln\frac1{\sqrt3}\\
=&\,\boxed{4-\frac32\ln3}.
\end{align*} I = = = 2 3 ( 3 − 3 1 ) + 3 ln ( 3 /2 1/2 ) 4 + 3 ln 3 1 4 − 2 3 ln 3 .
解法二
思路
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也可以在原变量下先做多项式除法。分母的导数与余下分子的 x x x
答题过程
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Working directly with the original integral,
x 3 3 + x 2 = x − 3 x 3 + x 2 . \frac{x^3}{3+x^2}=x-\frac{3x}{3+x^2}. 3 + x 2 x 3 = x − 3 + x 2 3 x . Consequently,
I = ∫ 1 3 ( x − 3 x 3 + x 2 ) d x = [ x 2 2 − 3 2 ln ( 3 + x 2 ) ] 1 3 = 9 − 1 2 − 3 2 ( ln 12 − ln 4 ) = 4 − 3 2 ln 3 . \begin{align*}
I=&\,\int_1^3\bigg(x-\frac{3x}{3+x^2}\bigg)\,\mathrm{d}x\\
=&\,\bigg[\frac{x^2}{2}-\frac32\ln(3+x^2)\bigg]_1^3\\
=&\,\frac{9-1}{2}-\frac32(\ln12-\ln4)\\
=&\,\boxed{4-\frac32\ln3}.
\end{align*} I = = = = ∫ 1 3 ( x − 3 + x 2 3 x ) d x [ 2 x 2 − 2 3 ln ( 3 + x 2 ) ] 1 3 2 9 − 1 − 2 3 ( ln 12 − ln 4 ) 4 − 2 3 ln 3 . 