定积分的换元积分法

定积分的换元积分法

\[设函数f(x)在区间[a,b]上连续，函数x=\varphi(t)满足条件：\\ (1)\varphi(\alpha)=a,\varphi(\beta)=b\\ (2)\varphi(t)在[\alpha,\beta](或[\beta,\alpha])上具有连续导数，且值域R_{\varphi}=[a,b],则有\\ \int_a^bf(x)dx=\int_{\alpha}^{\beta}f([\varphi(t)])\varphi'(t)dt \]

\[证明：设F'(x)=f(x),则\int_a^bf(x)dx=F(b)-F(a)\\ 设\Phi(t)=F(\varphi(t)),则\Phi'(t)=F'(\varphi(t))\varphi'(t)=f(\varphi(t))\varphi'(t)\\ 从而\int_{\alpha}^{\beta}f[\varphi(t)]\varphi'(t)dt=\Phi(t)\Big|_{\alpha}^{\beta}=\Phi(\beta)-\Phi(\alpha)=F(\varphi(\beta))-F(\varphi(\alpha))=F(b)-F(a)\\ \therefore \int_a^bf(x)dx=\int_{\alpha}^{\beta}f[\varphi(t)]\varphi'(t)dt=_{dx=\varphi'(t)dt}^{令x=\varphi(t)}\int_{\alpha}^{\beta}f(\varphi(t))\cdot\varphi'(t)dt \]

\[【例】\int_0^{\pi}\sqrt{\sin ^3x-\sin^5x}dx\\ 解：=\int_0^{\pi}\sqrt{\sin^3x(1-\sin^2)}dx\\ =\int_0^\pi\sqrt{\sin ^3x\cos^2x}dx\\ =\int_0^\pi\sin^{\frac32}x\cdot|\cos x|dx\\ =\int_0^{\frac\pi2}\sin^{\frac32}x.\cos xdx-\int_{\frac\pi2}^\pi\sin^{\frac32}x\cdot\cos xdx=\frac45 \]

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[例]（积偶函数的积分性质）证明：

\[(1)若f(x)在[-a,a]上连续且为偶函数，则\int_{-a}^af(x)dx=2\int_0^af(x)dx\\ (2)若f(x)在[-a,a]上连续且为奇函数，则f_{-a}^af(x)dx=0 \]

\[证明：\\ \int_{-a}^af(x)dx=\int_{-a}^0f(x)dx+\int_0^af(x)dx\\ 又\int_{-a}^0f(x)dx=_{dt=-dx}^{令t=-x}-\int_a^0f(-t)dt=\int_0^af(-t)dt\\ 若f(x)为偶函数，则有f(-t)=f(t)\\ 从而\int_{-a}^0f(x)dx=\int_0^af(t)dt=\int_0^af(x)dx\\ 则有\int_{-a}^af(x)dx=2\int_0^af(x)dx\\ 若有f(x)为奇函数，则有f(-t)=f(t)\\ 从而\int_{-a}^0f(x)dx=\int_0^af(-t)dt=-\int_0^af(t)dt=-\int_0^af(x)dx \]

周期函数的积分性质

\[设f(x)是周期函数，周期为T证明：\\ (1)\int_{a}^{a+T}f(x)dx=\int_{0}^Tf(x)dx\\ (2)\int_a^{a+nT}f(x)dx=n\int_0^Tf(x)dx(n\in N) \]

\[证明：(1)\int_a^{a+T}f(x)dx=\int_a^0f(x)dx+\int_0^Tf(x)dx+\int_T^{a+T}\\ 又\int_T^{a+T}f(x)dx=^{令u=x-T}\int_0^af(u+T)du\\ =\int_0^af(u)du=\int_0^af(x)dx\\ 故\int_A^{a+T}f(x)dx=\int_0^Tf(x)dx\\ \int_a^{a+T}f(x)dx=\int_{-\frac T2}^{\frac T2}f(x)dx \]

积分区间再现公式

\[设函数f(x)在[a,b]上连续，证明\\ \int_a^bf(x)dx=\int_a^b(a+b-x)dx \]

\[证明：\int_a^bf(x)dx=_{dt=-dx}^{t=a+b-x}-\int_b^af(a+b-t)dt\\ =\int_a^bf(a+b-t)dt=\int_a^bf(a+b-x)dx \]

\[【例】求\int_0^1x(1-x)^4dx\\ 解：=_{x=1-t}^{令t=1-x}-\int_1^0(1-t)t^4dt=\int_0^1(1-t)t^4dt\\ =\int_0^1t^4-t^5dt\\ =\frac1{30} \]

\[【例】求\int_0^1x\sqrt{1-x}dx\\ 解：令t=1-x\\ -\int_1^0(1-t)\sqrt tdt\\ =\int_0^1(1-t)\sqrt tdt=\int_0^1\sqrt tdt-\int_0^1t^{\frac32}dt\\ =\frac4{15} \]

\[【例】计算\int_0^{n\pi}x|\sin x|dx\\ 令t=n\pi-x\\ -\int_{n\pi}^0(n\pi-t)|\sin(n\pi-t)|dt\\ \int_0^{n\pi}(n\pi-t)|\sin t|dt\\ =n\pi\int_0^{n\pi}|\sin t|dt-\int_0^{n\pi}t|\sin t|dt\\ \Rightarrow2\int_0^{n\pi}x|\sin x|dx=n\pi\int_0^{n\pi}|\sin t|dt\\ \Rightarrow\int_0^{n\pi}x|\sin x|dt=\frac{n\pi}2\cdot n\int_0^\pi\sin tdt=n^2\pi \]

\[【例】设f(x)在[0,1]上连续，证明\\ (1)\int_0^{\frac\pi2}f(\sin x)dx=\int_0^{\frac\pi2}f(\cos x)dx\\ (2)\int_0^\pi xf(\sin x)dx=\frac \pi2\int_0^\pi f(\sin x)dx,由此计算\int_0^\pi\frac{x\sin x}{1+\cos ^2x}dx \]

\[证明(1)\int_0^{\frac\pi2}f(\sin x)dx\qquad 令t=\frac\pi2-x\\ =-\int_{\frac\pi2}^0f[\sin (\frac\pi2-t)]dt\\ =\int_0^{\frac\pi2}f(\cos t)dt=\int_0^{\frac\pi2}f(\cos x)dx \]

\[(2)\int_0^{\pi}xf(\sin x)dx\qquad 令t=\pi-x\\ -\int_{\pi}^0(\pi-t)f(\sin (\pi-t))dt\\ =\pi\int_0^{\pi}f(\sin t)dt-\int_0^\pi tf(\sin t)dt=\pi\int_0^\pi f(\sin x)dx-\int_0^\pi xf(\sin x)dx\\ \Rightarrow2\int_0^\pi xf(\sin x)dx=\pi\int_0^\pi f(\sin x)dx\\ \Rightarrow \int_0^\pi xf(\sin x)dx=\frac \pi2 \int_0^\pi f(\sin x)dx \]

\[\int_0^\pi \frac{x\sin x}{1+\cos^2x}dx=\frac \pi2\int_0^\pi \frac {\sin x}{1+\cos^2x}dx\\ =-\frac\pi2\int_0^\pi\frac1{1+\cos^2x}d\cos x=\frac{\pi^2}4 \]