题目
Problem
Find the exact value of ∫ 1 4 π 1 3 π 3 sin x sin 2 x d x \displaystyle\int_{\frac14\pi}^{\frac13\pi}3\sin x\sin2x\,\mathrm{d}x ∫ 4 1 π 3 1 π 3 sin x sin 2 x d x
Give your answer in the form p 3 + q 2 p\sqrt3+q\sqrt2 p 3 + q 2 p p p q q q
(4)
题目中文翻译
求 ∫ 1 4 π 1 3 π 3 sin x sin 2 x d x \displaystyle\int_{\frac14\pi}^{\frac13\pi}3\sin x\sin2x\,\mathrm{d}x ∫ 4 1 π 3 1 π 3 sin x sin 2 x d x
答案写成 p 3 + q 2 p\sqrt3+q\sqrt2 p 3 + q 2 p p p q q q
(4)
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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用二倍角把被积函数化为 sin 2 x cos x \sin^2x\cos x sin 2 x cos x
答题过程
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Using sin 2 x = 2 sin x cos x \sin2x=2\sin x\cos x sin 2 x = 2 sin x cos x
I = ∫ π / 4 π / 3 6 sin 2 x cos x d x = [ 2 sin 3 x ] π / 4 π / 3 = 2 [ ( 3 2 ) 3 − ( 2 2 ) 3 ] = 3 4 3 − 1 2 2 . \begin{align*}
I=&\,\int_{\pi/4}^{\pi/3}6\sin^2x\cos x\,\mathrm{d}x\\
=&\,\big[2\sin^3x\big]_{\pi/4}^{\pi/3}\\
=&\,2\bigg[\bigg(\frac{\sqrt3}{2}\bigg)^3
-\bigg(\frac{\sqrt2}{2}\bigg)^3\bigg]\\
=&\,\boxed{\frac34\sqrt3-\frac12\sqrt2}.
\end{align*} I = = = = ∫ π /4 π /3 6 sin 2 x cos x d x [ 2 sin 3 x ] π /4 π /3 2 [ ( 2 3 ) 3 − ( 2 2 ) 3 ] 4 3 3 − 2 1 2 .
解法二
思路
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连续分部积分两次会重新出现原积分。把这些积分项移到同一边，即可求出定积分。
答题过程
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Let
I = ∫ π / 4 π / 3 3 sin x sin 2 x d x . I=\int_{\pi/4}^{\pi/3}3\sin x\sin2x\,\mathrm{d}x. I = ∫ π /4 π /3 3 sin x sin 2 x d x . Integrating by parts twice gives
I = [ − 3 cos x sin 2 x ] π / 4 π / 3 + ∫ π / 4 π / 3 6 cos x cos 2 x d x , \begin{align*}
I=&\,\big[-3\cos x\sin2x\big]_{\pi/4}^{\pi/3}\\[4mm]
&\,\hspace{2pt}+\int_{\pi/4}^{\pi/3}6\cos x\cos2x\,\mathrm{d}x,
\end{align*} I = [ − 3 cos x sin 2 x ] π /4 π /3 + ∫ π /4 π /3 6 cos x cos 2 x d x , where
∫ π / 4 π / 3 6 cos x cos 2 x d x = [ 6 sin x cos 2 x ] π / 4 π / 3 + ∫ π / 4 π / 3 12 sin x sin 2 x d x = [ 6 sin x cos 2 x ] π / 4 π / 3 + 4 I . \begin{align*}
&\,\int_{\pi/4}^{\pi/3}6\cos x\cos2x\,\mathrm{d}x\\
=&\,\big[6\sin x\cos2x\big]_{\pi/4}^{\pi/3}
+\int_{\pi/4}^{\pi/3}12\sin x\sin2x\,\mathrm{d}x\\
=&\,\big[6\sin x\cos2x\big]_{\pi/4}^{\pi/3}+4I.
\end{align*} = = ∫ π /4 π /3 6 cos x cos 2 x d x [ 6 sin x cos 2 x ] π /4 π /3 + ∫ π /4 π /3 12 sin x sin 2 x d x [ 6 sin x cos 2 x ] π /4 π /3 + 4 I . Consequently, moving 4 I 4I 4 I
− 3 I = [ − 3 cos x sin 2 x + 6 sin x cos 2 x ] π / 4 π / 3 . -3I=\big[-3\cos x\sin2x+6\sin x\cos2x\big]_{\pi/4}^{\pi/3}. − 3 I = [ − 3 cos x sin 2 x + 6 sin x cos 2 x ] π /4 π /3 . Dividing by − 3 -3 − 3
I = [ cos x sin 2 x − 2 sin x cos 2 x ] π / 4 π / 3 . I=\big[\cos x\sin2x-2\sin x\cos2x\big]_{\pi/4}^{\pi/3}. I = [ cos x sin 2 x − 2 sin x cos 2 x ] π /4 π /3 . At the upper limit this expression is 3 3 / 4 3\sqrt3/4 3 3 /4 2 / 2 \sqrt2/2 2 /2
I = 3 4 3 − 1 2 2 . \boxed{I=\frac34\sqrt3-\frac12\sqrt2}. I = 4 3 3 − 2 1 2 . 