题目
Problem
(a) Find
∫x2cos2xdx
(4)
(b) Hence solve the differential equation
dtdy=(ytcost)2
giving your answer in the form yn=f(t) where n is an integer.
(5)
题目中文翻译
(a) 求
∫x2cos2xdx
(4)
(b) 进而解微分方程
dtdy=(ytcost)2
并将答案写成 yn=f(t) 的形式,其中 n 为整数。
(5)
解答
(a)
解法一
思路
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使用两次分部积分:先对 x2 求导,再处理剩下的 xsin2x。把两个积分分别记为 I 和 K,有助于看清符号并保持每行较短。
答题过程
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Let
I=∫x2cos2xdx.
For integration by parts, take u=x2 and dv=cos2xdx. Then du=2xdx and v=21sin2x, so
I=21x2sin2x−K,
where
K=∫xsin2xdx.
For K, take u=x and dv=sin2xdx. Now du=dx and v=−21cos2x. Hence
K==−21xcos2x+21∫cos2xdx−21xcos2x+41sin2x.
Substituting back,
I=21x2sin2x+21xcos2x−41sin2x+C.
解法二
思路
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也可以用表格法组织分部积分。不断对 x2 求导、对 cos2x 积分,再取带交替符号的对角乘积。
答题过程
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The repeated integration by parts table is
Dx22x20Icos2x21sin2x−41cos2x−81sin2x
Take the three diagonal products with alternating signs:
x2(21sin2x)=21x2sin2x,
−2x(−41cos2x)=21xcos2x,
and
2(−81sin2x)=−41sin2x.
Therefore
∫x2cos2xdx=21x2sin2x+21xcos2x−41sin2x+C.
(b)
解法一
思路
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先分离变量,再用双角公式把右侧拆成两个积分。题目要求 Hence,因此明确调用 (a) 的原函数,并保留积分常数,最后写成 y3=f(t)。
答题过程
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Rearranging the differential equation,
y2dtdy=t2cos2t.
Separate variables and integrate:
∫y2dy=∫t2cos2tdt.
The left side is 31y3. On the right, use
cos2t=21(1+cos2t).
Let F(t) denote the antiderivative from part (a), with x replaced by t:
F(t)=21t2sin2t+21tcos2t−41sin2t.
Then
∫t2cos2tdt==21∫t2dt+21∫t2cos2tdt61t3+21F(t)+C.
Thus
31y3=61t3+21F(t)+C.
Multiply by 3 and absorb the constant:
y3=21t3+23F(t)+C.
Substituting F(t) gives the required form:
y3=21t3+43t2sin2t+43tcos2t−83sin2t+C.
Here n=3 and C is an arbitrary constant.