问题116: 计算f(x)=1/(4+2tan(x)) 的原函数以及定积分

解法一: 三角换元

$$ \begin{aligned} & \int \frac{1}{4+2 \tan x} d x \\ = & \int \frac{1}{4+2 t} \cdot \frac{1}{1+t^2} d t \quad(t=\tan x) \\ = & \frac{1}{2} \int \frac{1}{(t+2)\left(t^2+1\right)} d t \end{aligned} $$

$$ \begin{aligned} & \frac{1}{(t+2)\left(t^2+1\right)}=\frac{1+t^2-t^2-2 t+2 t+4-4}{(t+2)\left(t^2+1\right)}=\frac{1}{t+2}+\frac{2-t}{t^2+1}-\frac{4}{(t+2)\left(t^2+1\right)} \\ & \frac{1}{(t+2)\left(t^2+1\right)}=\frac{1}{5}\left(\frac{1}{t+2}+\frac{2-t}{t^2+1}\right) \end{aligned} $$

$$ \begin{aligned} \text { 原式 } & =\frac{1}{10} \int \frac{1}{t+2}+\frac{2}{t^2+1}-\frac{t}{t^2+1} d t \\ & =\frac{1}{10}\left(\ln |t+2|+2 \arctan t-\frac{1}{2} \int \frac{d t^2}{t^2+1}\right) \\ & =\frac{1}{10} \ln |t+2|+\frac{1}{5} \arctan t-\frac{1}{20} \ln \left(t^2+1\right)+C \\ & =\frac{1}{10} \ln |2+\tan x|+\frac{1}{5} x-\frac{1}{20} \ln \left(\tan ^2 x+1\right)+C \\ & =\frac{1}{10} \ln |2+t \cos x|+\frac{1}{10} \ln |\cos x|+\frac{1}{5} x+C \\ & =\frac{1}{10} \ln |2 \cos x+\sin x|+\frac{1}{5} x+C \end{aligned} $$

$$ \begin{aligned} \int_0^{\frac{\pi}{4}} f(x) d x & =\frac{1}{10} \ln \left|2 \cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\right|+\frac{1}{5} \cdot \frac{\pi}{4}-\frac{1}{10} \ln 2 \\ & =\frac{1}{16} \ln \frac{3 \sqrt{2}}{4}+\frac{\pi}{20} \\ & =\frac{1}{20}\left(\pi+\ln \frac{9}{8}\right) \end{aligned} $$

解法二: 凑方法

$$ \begin{aligned} & \int \frac{1}{4+2 \tan x} d x \\ = & \frac{1}{2} \int \frac{\cos x}{2 \cos x+\sin x} d x \\ = & \frac{1}{2} \int \frac{\cos x-2 \sin x+2 \sin x}{2 \cos x+\sin x} d x \\ = & \frac{1}{2} \int \frac{d(2 \cos x+\sin x)}{2 \cos x+\sin x}+\int \frac{\sin x+2 \cos x-2 \cos x}{2 \cos x+\sin x} d x \\ = & \frac{1}{2} \ln |2 \cos x+\sin x|+\int 1 d x-2 \int \frac{\cos x}{2 \cos x+\sin x} d x \\ = & \frac{1}{5}\left(\frac{1}{2} \ln |2 \cos x+\sin x|+x\right)+C \\ = & \frac{1}{10} \ln |2 \cos x+\sin x|+\frac{1}{5} x+C \end{aligned} $$

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最后修改于1月23日