换元积分法训练题
在求解不定积分的过程中，第一和第二换元积分法的应用不是彼此孤立的，往往需要同时混合使用
instance 0
\[\begin{align}
\int x^{3}\sqrt{4-x^{2}}dx=?
\\ \\
设:x=2\sin t
\\ \\
\int\left(2\sin t\right)^{3}\sqrt{4-4\sin^{2}t} \cdot d\left(2\sin t\right)
\\ \\
\int(2\sin t)^{3}\sqrt{4(1-\sin^{2}t)}2\cos tdt
\\ \\
\int8\sin^{3}t\cdot\sqrt{4\cos^{2}t}\cdot2\cos tdt
\\ \\
\int8\sin^{3}t\cdot2\cos t\cdot2\cos tdt
\\ \\
32\int\sin^{3}t\cos^{2}tdt
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设:u=\cos t
\\ \\
du=d(\cos t)=(\cos t)^{\prime}dt=-\sin tdt
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\frac{du}{-\sin t}=dt
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32\int\sin^{3}t u^{2}\cdot\frac{du}{-\sin t}
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-32\int\sin^{2}t u^{2}du
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-32\int(1-u^{2})u^{2}du=-32\int(u^{2}-u^{4})du
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=-32\left(\frac{1}{3}u^{3}-\frac{1}{5}u^{5}\right)+C
\\ \\
=-\frac{32}{3}\cos^{3}t+\frac{32}{5}\cos^{5}t+C
\\ \\
\because\sin t=\frac{x}{2}
\\ \\
根据勾股定理: \cos t=\frac{\sqrt{4-x^{2}}}{2}
\\ \\
\cos^{3}t=(\frac{\sqrt{4-x^{2}}}{2})^{3}=\frac{(4-x^{2})^{\frac{3}{2}}}{8}
\\ \\
\cos^{5}t=(\frac{\sqrt{4-x^{2}}}{2})^{5}=\frac{(4-x^{2})^{\frac{5}{2}}}{32}
\\ \\
-\frac{32}{3}\cdot\frac{(4-x^{2})^{\frac{3}{2}}}{8}+\frac{32}{5}\cdot\frac{(4-x^{2})^{\frac{5}{2}}}{32}+C
\\ \\
=-\frac{4}{3}(4-x^{3})^{\frac{3}{2}}+\frac{1}{5}(4-x^{2})^{\frac{5}{2}}+C
\end{align}
\]
paradigm 0
援引题Sample 0
\[\begin{align}
\int \sqrt{3+2 x-x^{2}} d x=?
\\ \\
\int \sqrt{3-x^{2}+2 x-1+1} d x
\\ \\
\int \sqrt{3+\left[-\left(x^{2}-2 x+1\right)\right]+1} d x
\\ \\
\int \sqrt{4+\left[-(x-1)^{2}\right]} d x
\\ \\
\text { 设: } \quad t=x-1
\\ \\
\Rightarrow \int \sqrt{4-t^{2}} d(x-1)=\int \sqrt{2^{2}-t^{2}} d t
\\ \\
根据援引题:\text {Sample 0}
\\ \\
\Rightarrow \frac{x-1}{2} \sqrt{2^{2}-(x-1)^{2}}+\frac{2^{2}}{2} \arcsin \frac{x-1}{2}+C
\\ \\
=\frac{x-1}{2} \sqrt{3+2 x-x^{2}}+2 \arcsin \frac{x-1}{2}+C
\end{align}
\]
paradigm 1
援引题: Sample3
\[\begin{align}
\quad \int \frac{2 x+1}{\sqrt{x^{2}+2 x+5}} d x=?
\\ \\
\Rightarrow \int \frac{2 x+1}{\sqrt{\left(x^{2}+2 x+1\right)+4}} d x=\int \frac{2 x+1}{\sqrt{(x+1)^{2}+2^{2}}} d x
\\ \\
\text { 设: } u=x+1
\\ \\
\text { 则: } x=u-1
\\ \\
\int \frac{2(u-1)+1}{\sqrt{u^{2}+2^{2}}} d(x+1)
\\ \\
\Rightarrow \int \frac{2 u-1}{\sqrt{u^{2}+2^{2}}} d u=\int \frac{2 u}{\sqrt{u^{2}+2^{2}}} d u-\int \frac{1}{\sqrt{u^{2}+2^{2}}} d u
\\ \\ \\
解决第一式:
\\
再设: u^{2}+2^{2}=h
\\ \\
\int \frac{2 u}{\sqrt{h}} d\left(u^{2}+2^{2}\right) \Rightarrow \int \frac{2 u}{\sqrt{h}} \cdot \frac{d h}{2 u}=\int \frac{1}{\sqrt{h}} d h
\\ \\
\int \frac{1}{\sqrt{h}} d h=\int \frac{1}{h^{\frac{1}{2}}} d h=
\int h^{-\frac{1}{2}} d h=
\left[1 \div (-\frac{1}{2}+1) \right] h^{-\frac{1}{2}+1}+C
\\ \\
=2 h^{\frac{1}{2}}+C
\\ \\ \\
\text { 解决第二式, 根据援引题Sample } 3:
\\ \\
\int \frac{1}{\sqrt{u^{2}+2^{2}}} d u=
\ln \left(\sqrt{u^{2}+2^{2}}+u\right)+C
\\ \\ \\
2 h^{\frac{1}{2}} - \ln \left(\sqrt{u^{2}+2^{2}}+u\right) +C
\\ \\
=2(x^{2}+2x+5)^{\frac{1}{2}}-\ln (\sqrt[]{x^{2}+2x+5}+x+1)+C
\end{align}
\]
