问题及解答

求 $I_n=\displaystyle\int_{0}^{\frac{\pi}{4}}\tan^n\theta\mathrm{d}\theta$.

特别的, 

\[
I_1=\int_{0}^{\frac{\pi}{4}}\tan\theta\mathrm{d}\theta=\frac{1}{2}\ln 2,
\]

\[
I_2=\int_{0}^{\frac{\pi}{4}}\tan^2\theta\mathrm{d}\theta=1-\frac{\pi}{4}.
\]

当 $n\geqslant 2$ 时, 有递推公式

\[
I_n=\frac{1}{n-1}-I_{n-2}.
\]

\[
I_{2n}=(-1)^n\biggl[\frac{\pi}{4}-(1-\frac{1}{3}+\frac{1}{5}-\cdots+\frac{(-1)^{n-1}}{2n-1})\biggr]
\]

这里 $I_{2n}$ 参见 [1]

对于 $\tan^n x$ 的不定积分, 也有类似的递推公式. 

若记 $J_n=\int\tan^n x\mathrm{d}x$, ($n\geqslant 2$, 且 $n$ 是整数.) 则有

\[
J_n=\frac{1}{n-1}\tan^{n-1}x-J_{n-2} .
\]

References:

[1] 吉米多维奇, 问题2283

\[
I_1=\int_{0}^{\frac{\pi}{4}}\tan\theta\mathrm{d}\theta
=\int_{0}^{\frac{\pi}{4}}\frac{\sin\theta}{\cos\theta}\mathrm{d}\theta
=\int_{0}^{\frac{\pi}{4}}\frac{-1}{\cos\theta}\mathrm{d}\cos\theta
\]

令 $x=\cos\theta$, 则

\[
\begin{split}
\int_{0}^{\frac{\pi}{4}}\frac{-1}{\cos\theta}\mathrm{d}\cos\theta&=\int_{1}^{\frac{\sqrt{2}}{2}}\frac{-1}{x}\mathrm{d}x\\
&=\int_{\frac{\sqrt{2}}{2}}^{1}\frac{1}{x}\mathrm{d}x\\
&=(\ln x)\biggr|_{\frac{\sqrt{2}}{2}}^{1}\\
&=\frac{1}{2}\ln 2.
\end{split}
\]

\[
I_2=\int_{0}^{\frac{\pi}{4}}\tan^2\theta\mathrm{d}\theta
=\int_{0}^{\frac{\pi}{4}}(\sec^2\theta-1)\mathrm{d}\theta
=(\tan\theta-\theta)\biggr|_{0}^{\frac{\pi}{4}}
=1-\frac{\pi}{4}.
\]

对于 $n\geqslant 2$, 

\[
\begin{split}
I_n&=\int_{0}^{\frac{\pi}{4}}\tan^n\theta\mathrm{d}\theta\\
&=\int_{0}^{\frac{\pi}{4}}\tan^{n-2}\theta\cdot\tan^2\theta\mathrm{d}\theta\\
&=\int_{0}^{\frac{\pi}{4}}\tan^{n-2}\theta\cdot(\sec^2\theta-1)\mathrm{d}\theta\\
&=\int_{0}^{\frac{\pi}{4}}\tan^{n-2}\theta\cdot\sec^2\theta\mathrm{d}\theta-\int_{0}^{\frac{\pi}{4}}\tan^{n-2}\theta\mathrm{d}\theta\\
&=\int_{0}^{\frac{\pi}{4}}\tan^{n-2}\theta\mathrm{d}\tan\theta-I_{n-2}\\
\end{split}
\]

其中第一个积分中令 $x=\tan\theta$, 则

\[
\int_{0}^{\frac{\pi}{4}}\tan^{n-2}\theta\mathrm{d}\tan\theta=\int_0^1 x^{n-2}\mathrm{d}x=\frac{1}{n-1}x^{n-1}\biggr|_0^1=\frac{1}{n-1}.
\]

因此得到递推公式

\[
I_n=\frac{1}{n-1}-I_{n-2}.
\]