依据不定积分的换元积分法和分部积分法，给出定积分相对应的积分方法
一、换元积分法
\int _{a}^{b} f(x)\, dx =\int _{\alpha}^{\beta} f\left[\varphi (t)\right]\varphi'(t) \, dt
\end{align}$$
注意换元要换积分限
$$\begin{align}
\int _{0}^{\pi/2} f(\sin x)\, dx & =-\int _{\pi/ 2}^{0} f\left[\sin(\dfrac{\pi}{2}-t)\right]\, dt=\int _{0}^{\pi/2} f(\cos t)\, dt \\
& = \int _{0}^{\pi/2} f(\cos x)\, dx
\end{align}$$
$$\begin{align}
\int _{0}^{\pi } xf(\sin x)\, dx & =-\int _{\pi }^{0} (\pi-t)f\left[\sin(\pi-t)\right]\, dt=\int _{0}^{\pi} (\pi-t)f(\sin t)\, dt \\
& =\pi \int _{0}^{\pi} f(\sin x) \, dx -\int _{0}^{\pi} xf(\sin x)\, dx \\
&=\dfrac{\pi}{2} \int _{0}^{\pi} f(\sin x) \, dx
\end{align}$$
- 如果 $f (x)$ 在区间 $[-a,a]$ 上连续，且为偶函数：
$$\begin{align}
\int _{-a}^{a} f(x)\, dx =2\int _{0}^{a} f(x)\, dx
\end{align}$$
- 如果 $f (x)$ 在区间 $[-a,a]$ 上连续，且为奇函数：
$$\begin{align}
\int _{-a}^{a} f(x)\, dx =0
\end{align}$$
- 如果 $f (x)$ 为连续的周期函数：
$$\begin{align}
\int _{a}^{a+T} f(x)\, dx =\int _{0}^{T} f(x)\, dx \\
\int _{a}^{a+nT} f(x)\, dx =n\int _{0}^{T} f(x)\, dx
\end{align}$$
### 二、分部积分法
$$\begin{align}
\int _{a}^{b} u\, dv=\left[uv\right]_{a}^{b}-\int _{a}^{b} v\, du
\end{align}$$
瓦里斯公式：点火公式
$$\begin{align}
I_{n} & =\int _{0}^{\pi/2} \sin ^{n}x\, dx =\int _{0}^{\pi/2} \cos ^{n}x\, dx \\ \\
& =\begin{cases}
\dfrac{(n-1)!!}{n!!}\cdot \dfrac{\pi}{2} & =\dfrac{n-1}{n}\cdot \dfrac{n-3}{n-2}\cdots \dfrac{1}{2}\cdot \dfrac{\pi}{2} \quad \text{n为正偶数}\\ \\
\quad \dfrac{(n-1)!!}{n!!} & =\dfrac{n-1}{n}\cdot \dfrac{n-3}{n-2}\cdots \dfrac{4}{5}\cdot \dfrac{2}{3}\quad \text{n为正奇数}
\end{cases}
\end{align}$$
证明过程：
$$\begin{align}
I_{n} & =\int _{0}^{\pi/2} \sin ^{n}x\, dx=-\int _{0}^{\pi/2} \sin ^{n-1}x\, d(\cos x) \\
& =\left[-\cos x\sin ^{n-1}x\right]_{0}^{\pi / 2} +(n-1) \int _{0}^{\pi /2} \sin ^{n-2}x \cos ^{2}x\, dx \\
&=0+(n-1)\int _{0}^{\pi/2}\sin ^{n-2}x\; \mathrm{d}x -(n-1)\int _{0}^{\pi/2} \sin ^{n}x\, \; \mathrm{d}x \\
&=(n-1)I_{n-2}-(n-1)I_{n} \\
&=\dfrac{n-1}{n}I_{n-2}
\end{align}$$
$$\begin{cases}
I_{2m}=\dfrac{2m-1}{2m}\cdot \dfrac{2m-3}{2m-5}\cdots \dfrac{3}{4}\cdot \dfrac{1}{2} I_{0} \\ \\
I_{2m+1}=\dfrac{2m}{2m+1}\cdot \dfrac{2m-2}{2m-1}\cdots \dfrac{4}{5}\cdot \dfrac{2}{3} I_{1}
\end{cases}$$
$$\begin{align}
I_{0}=\int _{0}^{\pi/2} \, dx =\dfrac{\pi}{2}
\end{align}$$
$$\begin{align}
I_{1}=\int _{0}^{\pi/2} \sin x\, dx =1
\end{align}$$
