在求解不定積分的過程中，第一和第二換元積分法的應用不是彼此孤立的，往往需要同時混合使用
instance 0
\[\begin{align}
\int x^{3}\sqrt{4-x^{2}}dx=?
\\ \\
設:x=2\sin t
\\ \\
\int\left(2\sin t\right)^{3}\sqrt{4-4\sin^{2}t} \cdot d\left(2\sin t\right)
\\ \\
\int(2\sin t)^{3}\sqrt{4(1-\sin^{2}t)}2\cos tdt
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\int8\sin^{3}t\cdot\sqrt{4\cos^{2}t}\cdot2\cos tdt
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\int8\sin^{3}t\cdot2\cos2t\cdot2\cos tdt
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32\int\sin^{3}t\cos^{2}tdt
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設:u=\cos t
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du=d(\cos t)=(\cos t)^{\prime}dt=-\sin tdt
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\frac{du}{-\sin t}=dt
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32\int\sin^{3}u^{2}\cdot\frac{du}{-\sin t}
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-32\int\sin^{2}u^{2}du
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-32\int(1-u^{2})u^{2}du=-32\int(u^{2}-u^{4})du
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=-32\left(\frac{1}{3}u^{3}-\frac{1}{5}u^{5}\right)+C
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=-\frac{32}{3}\cos^{3}t+\frac{32}{5}\cos^{5}t+C
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\because\sin t=\frac{x}{2}
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根據勾股定理: \cos t=\frac{\sqrt{4-x^{2}}}{2}
\\ \\
\cos^{3}t=(\frac{\sqrt{4-x^{2}}}{2})^{3}=\frac{(4-x^{2})^{\frac{3}{2}}}{8}
\\ \\
\cos^{5}t=(\frac{\sqrt{4-x^{2}}}{2})^{5}=\frac{(4-x^{2})^{\frac{5}{2}}}{32}
\\ \\
-\frac{32}{3}\cdot\frac{(4-x^{2})^{\frac{3}{2}}}{8}+\frac{32}{5}\cdot\frac{(4-x^{2})^{\frac{5}{2}}}{32}+C
\\ \\
=-\frac{4}{3}(4-x^{3})^{\frac{3}{2}}+\frac{1}{5}(4-x^{2})^{\frac{5}{2}}+C
\end{align}
\]