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
\[\begin{eqnarray} \int_{0}^{1}e^{\sqrt{x}}dx=? \\ \\ 设: \sqrt{x}=t, \enspace x=t^{2} \\ \\ \int_{0}^{1} e^{t}d(t^{2}) \\ \\ d(t^{2})=dt\cdot(t^{2})^{\prime}=dt\cdot2t \\ \\ \int_{0}^{1} e^{t}2tdt=2\int_{0}^{1} e^{t}tdt \\ \\ \Rightarrow2\int_{0}^{1}\left(e^{t}\right)^{\prime}tdt=2\int_{0}^{1}td\left(e^{t}\right) \\ \\ =2\left[te^{t}\right]_{0}^{1}-2\int_{0}^{1}e^{t}dt \\ \\ 2\left[te^{t}\right]_{0}^{1} =2\left[\sqrt{x}e^{\sqrt{x}}\right]_{0}^{1}=2\sqrt{1}e^{\sqrt{1}}-0=2e \\ \\ \\ \\ {2\int_{0}^{1}e^{t}dt=2\left[e^{t}\right]_{0}^{1} =2\left[e^{\sqrt{x}}\right]_{0}^{1}} \\ \\ 2\left[e^{\sqrt{x}}\right]_{0}^{1} =2e^{\sqrt{1}}-2e^{\sqrt{0}}=2e-2 \\ \\ 2e-(2e-2)=2 \end{eqnarray} \]标签：right,frac,int,积分,sqrt,积分法,分部,pi,left From： https://www.cnblogs.com/Preparing/p/18301735