题目
Problem
(a) Given that 2 x = tan y 2x=\tan y 2 x = tan y d y d x = 2 1 + 4 x 2 \frac{\mathrm{d}y}{\mathrm{d}x}=\frac2{1+4x^2} d x d y = 1 + 4 x 2 2
(3)
(b) Hence find the exact value of ∫ 1 2 3 2 x tan − 1 ( 2 x ) d x \displaystyle\int_{\frac12}^{\frac{\sqrt3}2}x\tan^{-1}(2x)\,\mathrm{d}x ∫ 2 1 2 3 x tan − 1 ( 2 x ) d x
(7)
题目中文翻译
(a) 已知 2 x = tan y 2x=\tan y 2 x = tan y d y d x = 2 1 + 4 x 2 \frac{\mathrm{d}y}{\mathrm{d}x}=\frac2{1+4x^2} d x d y = 1 + 4 x 2 2
(3)
(b) 由此求 ∫ 1 2 3 2 x tan − 1 ( 2 x ) d x \displaystyle\int_{\frac12}^{\frac{\sqrt3}2}x\tan^{-1}(2x)\,\mathrm{d}x ∫ 2 1 2 3 x tan − 1 ( 2 x ) d x
(7)
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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隐式求导正切关系，再把正割平方写成正切平方加一。
答题过程
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Differentiating 2 x = tan y 2x=\tan y 2 x = tan y x x x
2 = sec 2 y d y d x . 2=\sec^2y\frac{\mathrm{d}y}{\mathrm{d}x}. 2 = sec 2 y d x d y . Since sec 2 y = 1 + tan 2 y = 1 + 4 x 2 \sec^2y=1+\tan^2y=1+4x^2 sec 2 y = 1 + tan 2 y = 1 + 4 x 2
d y d x = 2 1 + 4 x 2 . \boxed{\frac{\mathrm{d}y}{\mathrm{d}x}
=\frac2{1+4x^2}}. d x d y = 1 + 4 x 2 2 .
(b)
解法一
思路
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承接(a)取得反正切导数，分部积分后把有理分式先除开，再求积分。
答题过程
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Using (a), the derivative of tan − 1 ( 2 x ) \tan^{-1}(2x) tan − 1 ( 2 x ) 2 / ( 1 + 4 x 2 ) 2/(1+4x^2) 2/ ( 1 + 4 x 2 )
I = ∫ 1 / 2 3 / 2 x tan − 1 ( 2 x ) d x . I=\int_{1/2}^{\sqrt3/2}x\tan^{-1}(2x)\,\mathrm{d}x. I = ∫ 1/2 3 /2 x tan − 1 ( 2 x ) d x . For integration by parts, take u = tan − 1 ( 2 x ) u=\tan^{-1}(2x) u = tan − 1 ( 2 x ) d v = x d x \mathrm{d}v=x\,\mathrm{d}x d v = x d x v = x 2 / 2 v=x^2/2 v = x 2 /2
I = [ x 2 2 tan − 1 ( 2 x ) ] 1 / 2 3 / 2 − ∫ 1 / 2 3 / 2 x 2 1 + 4 x 2 d x . \begin{align*}
I=&\,\bigg[\frac{x^2}{2}\tan^{-1}(2x)\bigg]_{1/2}^{\sqrt3/2}\\[4mm]
&\,\hspace{2pt}-\int_{1/2}^{\sqrt3/2}\frac{x^2}{1+4x^2}\,\mathrm{d}x.
\end{align*} I = [ 2 x 2 tan − 1 ( 2 x ) ] 1/2 3 /2 − ∫ 1/2 3 /2 1 + 4 x 2 x 2 d x . Since
x 2 1 + 4 x 2 = 1 4 − 1 4 ( 1 + 4 x 2 ) , \frac{x^2}{1+4x^2}=\frac14-\frac1{4(1+4x^2)}, 1 + 4 x 2 x 2 = 4 1 − 4 ( 1 + 4 x 2 ) 1 , an antiderivative is
F ( x ) = x 2 2 tan − 1 ( 2 x ) − x 4 + 1 8 tan − 1 ( 2 x ) . \begin{align*}
F(x)=&\,\frac{x^2}{2}\tan^{-1}(2x)-\frac x4\\[4mm]
&\,\hspace{2pt}+\frac18\tan^{-1}(2x).
\end{align*} F ( x ) = 2 x 2 tan − 1 ( 2 x ) − 4 x + 8 1 tan − 1 ( 2 x ) . At the limits,
F ( 3 2 ) = π 6 − 3 8 , F ( 1 2 ) = π 16 − 1 8 . \begin{align*}
F\bigg(\frac{\sqrt3}{2}\bigg)
=&\,\frac\pi6-\frac{\sqrt3}{8},\\[4mm]
F\bigg(\frac12\bigg)=&\,\frac\pi{16}-\frac18.
\end{align*} F ( 2 3 ) = F ( 2 1 ) = 6 π − 8 3 , 16 π − 8 1 . Therefore I = F ( 3 / 2 ) − F ( 1 / 2 ) I=F(\sqrt3/2)-F(1/2) I = F ( 3 /2 ) − F ( 1/2 )
I = 5 π 48 + 1 − 3 8 . \boxed{I=\frac{5\pi}{48}+\frac{1-\sqrt3}{8}}. I = 48 5 π + 8 1 − 3 . 