计算定积分 $\int_0^a \frac{\mathrm{d} x}{x+\sqrt{a^2-x^2}}$.
【答案】 【参考解析】设 $x=a \sin t$ ,则
$$
\begin{gathered}
\int_0^a \frac{\mathrm{d} x}{x+\sqrt{a^2-x^2}}=\int_0^{\frac{\pi}{2}} \frac{a \cos t \mathrm{dt}}{a \sin t+a \cos t} \\
=\int_0^{\frac{\pi}{2}} \frac{\cos t \mathrm{dt}}{\sin t+\cos t} \\
\text { 令 } I_1=\int_0^{\frac{\pi}{2}} \frac{\cos t \mathrm{~d}}{\sin t+\cos t}, I_2=\int_0^{\frac{\pi}{2}} \frac{\sin t \mathrm{dt}}{\sin t+\cos t} \text { ,则 } \\
I_1-I_2=\int_0^{\frac{\pi}{2}} \frac{\cos t-\sin t}{\sin t+\cos t} \mathrm{~d} t \\
=\int_0^{\frac{\pi}{2}} \frac{1}{\sin t+\cos t} \mathrm{~d}(\sin t+\cos t) \\
=[\ln (\sin t+\cos t)]_0^{\frac{\pi}{2}} \mathrm{~d} t=\frac{\pi}{2},
\end{gathered}
$$

$$
\begin{aligned}
&I_1-I_2=\int_0^{\frac{\pi}{2}} \frac{\cos t-\sin t}{\sin t+\cos t} \mathrm{~d} t \\
&=\int_0^{\frac{\pi}{2}} \frac{1}{\sin t+\cos t} \mathrm{~d}(\sin t+\cos t)
\end{aligned}
$$

$$
=[\ln (\sin t+\cos t)]_0^{\frac{\pi}{2}}=0
$$
所以 $I_1=\frac{\pi}{4}=\int_0^a \frac{\mathrm{d} x}{x+\sqrt{a^2-x^2}}$.


系统推荐