问题及解答

计算定积分

\[\int_{0}^{\frac{\pi}{4}}\frac{1-\sin 2x}{1+\sin 2x}\mathrm{d}x\]

[Hint]

利用 $\int_0^a f(x)\mathrm{d}x=\int_0^a f(a-x)\mathrm{d}x$.

首先, 通过换元($t=a-x$)容易可以证明 $\int_0^a f(x)\mathrm{d}x=\int_0^a f(a-x)\mathrm{d}x$. 

于是,

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