定积分之分部积分法
brief
设函数 \(u=u(x)\) 与 \(q=q(x)\) 在 \([a,b]\)上分别具有连续的导数: \(u'(x)\) 与 \(q'(x)\) , 则有分部定积分公式:
\[\int_{a}^{b} udq=[uq]_{a}^{b}-\int_{a}^{b} qdu
\]
instance 0
\[\begin{align}
\int_{0}^{\frac{\pi}{2}} x \cos (x) d x=?
\\ \\
\Rightarrow \int_{0}^{\frac{\pi}{2}} x(\sin x)^{\prime} d x \Rightarrow \int_{0}^{\frac{\pi}{2}} x d(\sin x)
\\ \\
\Rightarrow \left[x\sin x\right]_{0}^{\frac{\pi}{2}}
-\int_{0}^{\frac{\pi}{2}}\sin xdx
\\ \\
{[x \sin (x)]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2} \cdot \sin \left(\frac{\pi}{2}\right)-0=\frac{\pi}{2}}
\\ \\
\int_{0}^{\frac{\pi}{2}}\sin x d x=[-\cos (x)]_{0}^{\frac{\pi}{2}}
\\ \\
-\cos \left(\frac{\pi}{2}\right)=0, \quad -\cos (0)=-1
\\ \\
\Rightarrow [-\cos (x)]_{0}^{\frac{\pi}{2}}=0-(-1)=1
\\ \\
\Rightarrow\frac{\pi}{2}-1
\end{align}
\]
instance 1
\[\begin{eqnarray}
\int_{0}^{1}e^{\sqrt{x}}dx=?
\\ \\
设: \sqrt{x}=t, \enspace x=t^{2}
\\ \\
\int_{0}^{1} e^{t}d(t^{2})
\\ \\
d(t^{2})=dt\cdot(t^{2})^{\prime}=dt\cdot2t
\\ \\
\int_{0}^{1} e^{t}2tdt=2\int_{0}^{1} e^{t}tdt
\\ \\
\Rightarrow2\int_{0}^{1}\left(e^{t}\right)^{\prime}tdt=2\int_{0}^{1}td\left(e^{t}\right)
\\ \\
=2\left[te^{t}\right]_{0}^{1}-2\int_{0}^{1}e^{t}dt
\\ \\
2\left[te^{t}\right]_{0}^{1}
=2\left[\sqrt{x}e^{\sqrt{x}}\right]_{0}^{1}=2\sqrt{1}e^{\sqrt{1}}-0=2e
\\ \\ \\ \\
{2\int_{0}^{1}e^{t}dt=2\left[e^{t}\right]_{0}^{1}
=2\left[e^{\sqrt{x}}\right]_{0}^{1}}
\\ \\
2\left[e^{\sqrt{x}}\right]_{0}^{1}
=2e^{\sqrt{1}}-2e^{\sqrt{0}}=2e-2
\\ \\
2e-(2e-2)=2
\end{eqnarray}
\]
Instance 2
Part 0
\[\begin{align}
\int_{0}^{\frac{\pi}{2}}e^{2x}\cos xdx=?
\\ \\
\int_{0}^{\frac{\pi}{2}}e^{2x}(\sin x)^{'}dx\Rightarrow[e^{2x}\sin x]_{0}^{\frac{\pi}{2}}
-\int_{0}^{\frac{\pi}{2}}\sin xd(e^{2x})
\\ \\
延展:\int_{0}^{\frac{\pi}{2}}\sin d (e^{2x})
\\ \\
(e^{2x})^{\prime}=(e^{2x})^{\prime}\cdot(2x)^{\prime}=2e^{2x}
\\ \\
\Rightarrow2\int_{0}^{\frac{\pi}{2}}\sin xe^{2x}dx
\\ \\
延展:2\int_{0}^{\frac{\pi}{2}}\sin xe^{2x}dx
=2\int_{0}^{\frac{\pi}{2}}(-\cos x)' e^{2x} dx
\\ \\
=2[-\cos xe^{2x}]_{0}^{\frac{\pi}{2}}-2\int_{0}^{\frac{\pi}{2}}(-\cos x)d(e^{2x})
\\ \\
延展:2\int_{0}^{\frac{\pi}{2}}(-\cos x) d(e^{2x})
\\ \\
=-2\int_{0}^{\frac{\pi}{2}}\cos x\cdot2e^{2x}dx
\\ \\
=-4\int_{0}^{\frac{\pi}{2}}\cos xe^{2x}dx
\\ \\
\Rightarrow2[-\cos xe^{2\pi}]_{0}^{\frac{\pi}{2}}-(-4\int_{0}^{\frac{\pi}{2}}\cos xe^{2x}dx)
\end{align}
\]
Part 1
\[\begin{align}
设:I=\int_{0}^{\frac{\pi}{2}}e^{2x}\cos xdx
\\ \\
\int_{0}^{\frac{\pi}{2}}e^{2x}\cos xdx=[e^{2x}\sin x]_{0}^{\frac{\pi}{2}}
-
\int_{0}^{\frac{\pi}{2}}\sin d\left(e^{2x}\right)
\\ \\
=[e^{2x}\sin x]_{0}^{\frac{\pi}{2}}-(2[-\cos xe^{2x}]_{0}^{\frac{\pi}{2}}
+4\int_{0}^{\frac{\pi}{2}}\cos xe^{2x}dx)
\\ \\
I=[e^{2x}\sin x]_{0}^{\frac{\pi}{2}}-(2[-\cos xe^{2x}]_{0}^{\frac{\pi}{2}}
+4I)
\\ \\
I=e^{\pi}-2-4I \Rightarrow
5I=e^{\pi}-2
\\ \\
I=\frac{1}{5}(e^{\pi}-2)
\end{align}
\]
