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
\[\begin{align} \int x^{3}\sqrt{4-x^{2}}dx=? \\ \\ 设:x=2\sin t \\ \\ \int\left(2\sin t\right)^{3}\sqrt{4-4\sin^{2}t} \cdot d\left(2\sin t\right) \\ \\ \int(2\sin t)^{3}\sqrt{4(1-\sin^{2}t)}2\cos tdt \\ \\ \int8\sin^{3}t\cdot\sqrt{4\cos^{2}t}\cdot2\cos tdt \\ \\ \int8\sin^{3}t\cdot2\cos2t\cdot2\cos tdt \\ \\ 32\int\sin^{3}t\cos^{2}tdt \\ \\ 设:u=\cos t \\ \\ du=d(\cos t)=(\cos t)^{\prime}dt=-\sin tdt \\ \\ \frac{du}{-\sin t}=dt \\ \\ 32\int\sin^{3}u^{2}\cdot\frac{du}{-\sin t} \\ \\ -32\int\sin^{2}u^{2}du \\ \\ -32\int(1-u^{2})u^{2}du=-32\int(u^{2}-u^{4})du \\ \\ =-32\left(\frac{1}{3}u^{3}-\frac{1}{5}u^{5}\right)+C \\ \\ =-\frac{32}{3}\cos^{3}t+\frac{32}{5}\cos^{5}t+C \\ \\ \because\sin t=\frac{x}{2} \\ \\ 根据勾股定理: \cos t=\frac{\sqrt{4-x^{2}}}{2} \\ \\ \cos^{3}t=(\frac{\sqrt{4-x^{2}}}{2})^{3}=\frac{(4-x^{2})^{\frac{3}{2}}}{8} \\ \\ \cos^{5}t=(\frac{\sqrt{4-x^{2}}}{2})^{5}=\frac{(4-x^{2})^{\frac{5}{2}}}{32} \\ \\ -\frac{32}{3}\cdot\frac{(4-x^{2})^{\frac{3}{2}}}{8}+\frac{32}{5}\cdot\frac{(4-x^{2})^{\frac{5}{2}}}{32}+C \\ \\ =-\frac{4}{3}(4-x^{3})^{\frac{3}{2}}+\frac{1}{5}(4-x^{2})^{\frac{5}{2}}+C \end{align} \]标签：cos,frac,训练,换元,int,32,sqrt,积分法,sin From： https://www.cnblogs.com/Preparing/p/18198947