降次积分法

如在求${\displaystyle \int \cos ^{5}(x)\,dx\!}$时，需要先求得${\displaystyle \int \cos ^{n}(x)\,dx\!}$的降次公式，过程如下：

${\displaystyle I_{n}\,=\int \cos ^{n}(x)\,dx\!}$

${\displaystyle =\int \cos ^{n-1}(x)\cos(x)\,dx\!}$

${\displaystyle =\int \cos ^{n-1}(x)\,d(\sin(x))\!}$

${\displaystyle =\cos ^{n-1}(x)\sin(x)-\int \sin(x)\,d(cos^{n-1}(x))\!}$

${\displaystyle =\cos ^{n-1}(x)\sin(x)+(n-1)\int \sin(x)\cos ^{n-2}(x)\sin(x)\,dx\!}$

${\displaystyle =\cos ^{n-1}(x)\sin(x)+(n-1)\int \cos ^{n-2}(x)\sin ^{2}(x)\,dx\!}$

${\displaystyle =\cos ^{n-1}(x)\sin(x)+(n-1)\int \cos ^{n-2}(x)(1-\cos ^{2}(x))\,dx\!}$

${\displaystyle =\cos ^{n-1}(x)\sin(x)+(n-1)\int \cos ^{n-2}(x)\,dx-(n-1)\int \cos ^{n}(x)\,dx\!}$

${\displaystyle =\cos ^{n-1}(x)\sin(x)+(n-1)I_{n-2}-(n-1)I_{n}\,}$

${\displaystyle I_{n}+(n-1)I_{n}=\cos ^{n-1}(x)\sin(x)+(n-1)I_{n-2}\,}$

${\displaystyle nI_{n}=\cos ^{n-1}(x)\sin(x)+(n-1)I_{n-2}\,}$

${\displaystyle I_{n}={\frac {1}{n}}\cos ^{n-1}(x)\sin(x)+{\frac {n-1}{n}}I_{n-2}\,}$

因此${\displaystyle \int \cos ^{n}(x)\,dx\!}$可表示为：

${\displaystyle \int \cos ^{n}(x)\,dx={\frac {1}{n}}\cos ^{n-1}(x)\sin(x)+{\frac {n-1}{n}}\int \cos ^{n-2}(x)\,dx\!}$

将n=5代入，可得：

${\displaystyle n=5\,}$：${\displaystyle I_{5}={\tfrac {1}{5}}\cos ^{4}(x)\sin(x)+{\tfrac {4}{5}}I_{3}\,}$

${\displaystyle n=3\,}$：${\displaystyle I_{3}={\tfrac {1}{3}}\cos ^{2}(x)\sin(x)+{\tfrac {2}{3}}I_{1}\,}$

${\displaystyle \because I_{1}=\int \cos(x)\,dx=\sin(x)+C_{1}\,}$

${\displaystyle \therefore I_{3}={\tfrac {1}{3}}\cos ^{2}(x)\sin(x)+{\tfrac {2}{3}}\sin(x)+C_{2}\,}$，${\displaystyle C_{2}={\tfrac {2}{3}}C_{1}\,}$

${\displaystyle I_{5}={\frac {1}{5}}\cos ^{4}(x)\sin(x)+{\frac {4}{5}}\left[{\frac {1}{3}}\cos ^{2}(x)\sin(x)+{\frac {2}{3}}\sin(x)\right]+C\,}$，C为常数

除了上述的${\displaystyle \int \cos ^{n}(x)\,dx\!}$外，常见的降次公式还有：

${\displaystyle \int \sin ^{n}(x)\,dx=-{\frac {1}{n}}\sin ^{n-1}(x)\cos(x)+{\frac {n-1}{n}}\int \sin ^{n-2}(x)\,dx\!}$

${\displaystyle \int \tan ^{n}(x)\,dx={\frac {1}{n-1}}\tan ^{n-1}(x)-\int \tan ^{n-2}(x)\,dx\!}$

${\displaystyle \int (\ln(x))^{n}\,dx=x(\ln(x))^{n}-n\int (\ln(x))^{n-1}\,dx\!}$