タンジェントとそのn乗の不定積分

$\displaystyle\int \tan xdx\\ =\displaystyle\int \dfrac{\sin x}{\cos x}dx\\ =\displaystyle\int -\dfrac{(\cos x)'}{\cos x}dx\\ =-\log|\cos x|+C$

$\displaystyle\int \tan^2xdx\\ =\displaystyle\int \left(\dfrac{1}{\cos^2x}-1\right)dx\\ =\tan x-x+C$

$\displaystyle\int\tan^nxdx\\ =\displaystyle\int\left(\dfrac{1}{\cos^2x}-1\right)\tan^{n-2}xdx\\ =\displaystyle\int\dfrac{\tan^{n-2}x}{\cos^2x}dx-\int\tan^{n-2}xdx$

ここで，$(\tan x)'=\dfrac{1}{\cos^2x}$ に注意すると第一項は

$\dfrac{1}{n-1}\tan^{n-1}x$ と計算できる。

つまり，$\displaystyle\int \tan^n xdx=\dfrac{1}{n-1}\tan^{n-1}x-\int\tan^{n-2}xdx$

• $n$ が奇数 $(2k+1)$ のとき，

$\displaystyle\int \tan^{2k+1}xdx\\ =\displaystyle\dfrac{1}{2k}\tan^{2k}x-\int \tan^{2k-1}xdx\\ =\cdots\\ =\displaystyle\sum_{t=0}^{k-1}(-1)^t\dfrac{\tan^{2(k-t)}x}{2(k-t)}+(-1)^k\int\tan xdx$

$=\displaystyle\sum_{t=1}^{k}(-1)^{k-t}\dfrac{\tan^{2t}x}{2t}+(-1)^{k+1}\log|\cos x|+C$

• $n$ が偶数 $(2k)$ のとき，

$\displaystyle\int\tan^{2k}dx\\ =\displaystyle\dfrac{1}{2k-1}\tan^{2k-1}x-\int \tan^{2k-2}xdx\\ =\cdots\\ =\displaystyle\sum_{t=0}^{k-1}(-1)^t\dfrac{\tan^{2(k-t)-1}x}{2(k-t)-1}+(-1)^{k}\int 1dx$

$=\displaystyle\sum_{t=1}^k(-1)^{k-t}\dfrac{\tan^{2t-1}x}{2t-1}+(-1)^kx+C$