定积分之换元积分法公式
brief
设函数\(f(x)\)在区间\([a,b]\)上连续变化，函数\(x= \varphi (t)\)在区间\([\alpha,\beta]\)上具有连续的导数
当\(t\)在区间\([\alpha,\beta]\)上变化时，\(x= \varphi (t)\)的值在\([a,b]\)上变化
又 \(\varphi(\alpha)=a, \quad \varphi(\beta)=b\)
则:
\[\int_{a}^{b} f(x)dx=\int_{\alpha}^{\beta} f[\varphi(t)]\varphi '(t)dt
\]
prove
\[\begin{align}
已知条件:
\left\{\begin{array}{l}
x=\varphi(t), \quad\left(\alpha \leqslant t \leqslant \beta\right)
\\ \\
f(x), \quad(a \leqslant x \leqslant b)
\\ \\
\varphi(\alpha)=a, \quad \varphi(\beta)=b
\end{array}\right.
\\ \\ \\
根据牛顿莱布尼茨公式:
\\
\int_{d}^{b} f(x) d x=F(b)-F(a)
\\ \\
\text { 设: } \Phi(t)=F(x) =F[\varphi(t)] \enspace \text { (关键) }
\\ \\
\text { 链式法则: }
\\
\Phi^{\prime}(t)=(F[\varphi(t)])^{\prime}=F^{\prime}[\varphi(t)]
\varphi '(t)=f[\varphi(t)] \varphi^{\prime}(t)
\\ \\
\text { 等于: }
\\
\int f[\varphi(t)] \varphi^{\prime}(t) d t=\Phi(t)+C
\\ \\
\Rightarrow \int_{\alpha}^{\beta} f[\varphi(t)] \varphi^{\prime}(t) d t=
\Phi(\beta)-\Phi(\alpha)
\\ \\
\because \Phi(t)=F[\varphi(t)]
\\ \\
\Phi(\beta)-\Phi(\alpha)=F[\varphi(\beta)]-F[\varphi(\alpha)]
\\ \\
\because \varphi(\beta)=b, \quad \varphi(\alpha)=a
\\ \\
\Rightarrow F[\varphi(\beta)]-F[\varphi(\alpha)]=F(b)-F(a)
=\int_{d}^{b} f(x) d x
\\ \\
\therefore \int_{a}^{b} f(x) d x=\int_{\alpha}^{\beta} f[\varphi(t)] \varphi^{\prime}(t) d t
\end{align}
\]
example 0
Part 0
\[\begin{eqnarray}
\int_{0}^{a} \sqrt{a^{2}-x^{2}} d x, \quad(a>0)
\\
设: a \sin (t)=x
\\
先求: [\alpha, \beta]= ?
\\ \\
\because \varphi(\alpha)=A, \quad \varphi(\beta)=B
\\ \\
\because[A, B]=[0, a]
\\ \\
\therefore a \sin (t)=0, \enspace \sin (t)=0, \quad t=0=\alpha
\\ \\
\therefore a\sin(t)=a, \enspace \sin (t)=1, \quad t=\frac{\pi}{2}=\beta
\\ \\
{[\alpha, \beta]=\left[0, \frac{\pi}{2}\right]}
\\ \\ \\
\Rightarrow \int_{0}^{a} \sqrt{a^{2}-(a \sin t)^{2}} d(a \sin t)
\\ \\
条目1: \sqrt{a^{2}-(a \sin t)^{2}}=\sqrt{a^{2}\left(1-\sin ^{2} t\right)}=a \sqrt{\cos ^{2} t}=a \cos (t)
\\ \\
条目2: d(a \sin t)=d t \cdot(a \sin t)^{\prime}=d t \cdot a \cos t
\\ \\
\int_{0}^{\frac{\pi}{2}} a \cos (t) \cdot a \cos (t) d t=a^{2} \int_{0}^{\frac{\pi}{2}} \cos t \cdot \cos t d t
\\ \\
\\
\text { 根据三角函数积化和差公式: }
\\ \\
\Rightarrow a^{2} \int_{0}^{\frac{\pi}{2}} \frac{\cos (t+t)+\cos (t-t)}{2} d t=a^{2} \int_{0}^{\frac{\pi}{2}} \frac{\cos (2 t)+1}{2} d t
\\ \\
= \frac{a^{2}}{2} \int_{0}^{\frac{\pi}{2}} \cos (2 t) d t+\frac{a^{2}}{2} \int_{0}^{\frac{\pi}{2}} d t
\\ \\
\end{eqnarray}
\]
Part 1
\[\begin{eqnarray}
\text{条目1}:\frac{a^{2}}{2}\int_{0}^{\frac{\pi}{2}}dt=\frac{a^{2}}{2}[t]_{0}^{\frac{\pi}{2}}=\frac{a^{2}}{2}(\frac{\pi}{2})-0=\frac{a^{2}\pi}{4}
\\ \\
\text{条目2}:\int\cos(2t)dt\Rightarrow 设:u=2t
\\ \\
\int\cos(u)d(2t)\Rightarrow\int\cos(u)\frac{du}{2}\Rightarrow\frac{1}{2}\sin(u)+C
\\ \\
\Rightarrow\frac{1}{2}\sin(2t)+C
\\ \\
\therefore\frac{a^{2}}{2}\int_{0}^{\frac{\pi}{2}}\cos(2t)dt=\frac{1}{2}\cdot\frac{a^{2}}{2}\left[\sin(2t)\right]_{0}^{\frac{\pi}{2}}
\\ \\
=\frac{a^{2}}{4}\sin(2\cdot\frac{\pi}{2})-\frac{a^{2}}{4}\sin(0)=0
\\ \\
因此:\frac{a^{2}}{2}\int_{0}^{\frac{\pi}{2}}\cos(2t)dt+\frac{a^{2}}{2}\int_{0}^{\frac{\pi}{2}}dt=\frac{a^{2}\pi}{4}
\end{eqnarray}
\]
sample 1
part 0
\[\begin{align}
\int_{0}^{3}\frac{x}{\sqrt{1+x}}dx=?
\\ \\
\begin{cases}设:t=\sqrt{1+x}
\\ \\
则:x=t^{2}-1\end{cases}
\\ \\
\boxed{\begin{aligned}
\\
[a,b]=[0,3]
\\ \\
[\alpha,\beta]:\\
x=0\Rightarrow t^{2}-1=0\Rightarrow t=\pm1
\\ \\
\because\sqrt{1+x}>0,
\\
\therefore t=1=\alpha
\end{aligned}}
\\ \\
\begin{cases}
x=3\Rightarrow t^{2}-1=3\Rightarrow t=\pm2
\\ \\
\because\sqrt{1+x}>0,
\\
\therefore t=2=\beta
\\ \\
[\alpha,\beta]=[1,2]
\end{cases}
\\ \\ \\
\Rightarrow\int_{1}^{2}\frac{t^{2}-1}{t}d(t^{2}-1)
\\ \\
d(t^{2}-1)=dt(t^{2}-1)^{\prime}=dt\cdot2t
\\ \\
\Rightarrow\int_{1}^{2}\frac{t^{2}-1}{t}2tdt=2\int_{1}^{2}(t^{2}-1)dt
\end{align}
\]
part 1
\[\begin{aligned}
2\int_{1}^{2}t^{2}dt-2\int_{1}^{2}dt
\\ \\
2\int_{1}^{2}t^{2}dt=2[\frac{1}{1+2}t^{2+1}]_{1}^{2}=2[\frac{1}{3}t^{3}]_{1}^{2}
\\ \\
2\int_{1}^{2}dt=2[t]_{1}^{2}
\\ \\
\Rightarrow2[\frac{1}{3}t^{3}]^{2}-2[t]^{2}=2[\frac{1}{3}t^{3}-t]^{2}
\\ \\
\boxed{\Rightarrow 2[\frac{(\sqrt{1+x})^{3}}{3}-(\sqrt{1+x})]_{1}^{2}}
\\ \\
[\frac{(\sqrt{3})^{3}}{3}-\sqrt{3}]=0
\\ \\
[\frac{(\sqrt{2})^{3}}{3}-\sqrt{2}]=\frac{-\sqrt{2}}{3}
\\ \\
\Rightarrow2[0-(-\frac{\sqrt{2}}{3})]=\frac{2\sqrt{2}}{3}
\end{aligned}
\]
sample 2
援引题:example 9
\[\begin{align}
\int_{0}^{\frac{\pi}{2}}\cos^{5}x\sin xdx=?
\\ \\
设: \cos(x)=t
\\ \\
\Rightarrow\int_{0}^{\frac{\pi}{2}}\cos^{5}x(-\cos x)^{\prime}dx
\\ \\
(-\cos x)^{\prime}\cdot dx=d(-\cos x)=d(-t)
\\ \\
-(-\sin x)dx=d(-t)
\\ \\
\because dt=d(\cos x)=-\sin xdx
\\ \\
\therefore d(-t)=-dt
\\ \\
\Rightarrow\int_{0}^{\frac{\pi}{2}}t^{5}\cdot-dt
=-\int_{0}^{\frac{\pi}{2}}t^{5}dt=-[\frac{1}{1+5}t^{5+1}]_{0}^{\frac{\pi}{2}}
\\ \\
\Rightarrow-[\frac{1}{6}t^{6}]_{0}^{\frac{\pi}{2}}=-[\frac{1}{6}\cos^{6}(x)]_{0}^{\frac{\pi}{2}}
\\ \\
\Rightarrow-(\frac{1}{6}[\cos^{6}(\frac{\pi}{2})])-(\frac{1}{6}[\cos^{6}(0)])
\\ \\
\Rightarrow 0-(-\frac{1}{6}\cdot1)=\frac{1}{6}
\end{align}
\]
