同济高数 第10章 第10-5-4题

[AI解答]

[AI解答]

（1）
设
$$
I(a)=\int_{0}^{\frac{\pi}{2}} \ln\frac{1+a\cos x}{1-a\cos x}\cdot\frac{dx}{\cos x},\quad |a|<1.
$$
对参数 $a$ 求导（积分号下求导，满足一致收敛条件）：
$$
I'(a)=\int_{0}^{\frac{\pi}{2}} \frac{\partial}{\partial a}\left[\ln\frac{1+a\cos x}{1-a\cos x}\right]\frac{dx}{\cos x}.
$$
由于
$$
\frac{\partial}{\partial a}\ln\frac{1+a\cos x}{1-a\cos x}
= \frac{\cos x}{1+a\cos x} - \frac{-\cos x}{1-a\cos x}
= \frac{\cos x}{1+a\cos x} + \frac{\cos x}{1-a\cos x}
= \frac{2\cos x}{1-a^{2}\cos^{2}x}.
$$
因此
$$
I'(a)=\int_{0}^{\frac{\pi}{2}} \frac{2\cos x}{1-a^{2}\cos^{2}x}\cdot\frac{dx}{\cos x}
=2\int_{0}^{\frac{\pi}{2}} \frac{dx}{1-a^{2}\cos^{2}x}.
$$
利用恒等式 $\cos^{2}x = \frac{1+\cos 2x}{2}$，
$$
1-a^{2}\cos^{2}x = 1-\frac{a^{2}}{2}(1+\cos 2x)
= 1-\frac{a^{2}}{2} - \frac{a^{2}}{2}\cos 2x
= \frac{2-a^{2}}{2} - \frac{a^{2}}{2}\cos 2x.
$$
于是
$$
I'(a)=2\int_{0}^{\frac{\pi}{2}} \frac{dx}{\frac{2-a^{2}}{2} - \frac{a^{2}}{2}\cos 2x}
=4\int_{0}^{\frac{\pi}{2}} \frac{dx}{2-a^{2} - a^{2}\cos 2x}.
$$
令 $t=2x$，则 $dx = dt/2$，积分限 $0\to\pi$，
$$
I'(a)=4\int_{0}^{\pi} \frac{dt/2}{2-a^{2} - a^{2}\cos t}
=2\int_{0}^{\pi} \frac{dt}{2-a^{2} - a^{2}\cos t}.
$$
利用万能公式：令 $u=\tan\frac{t}{2}$，则 $\cos t = \frac{1-u^{2}}{1+u^{2}}$，$dt = \frac{2}{1+u^{2}}du$，
分母：
$$
2-a^{2} - a^{2}\frac{1-u^{2}}{1+u^{2}}
= \frac{(2-a^{2})(1+u^{2}) - a^{2}(1-u^{2})}{1+u^{2}}
= \frac{2-a^{2} + (2-a^{2})u^{2} - a^{2} + a^{2}u^{2}}{1+u^{2}}
= \frac{2-2a^{2} + 2u^{2}}{1+u^{2}}
= \frac{2(1-a^{2}+u^{2})}{1+u^{2}}.
$$
因此
$$
I'(a)=2\int_{0}^{\infty} \frac{\frac{2}{1+u^{2}} du}{\frac{2(1-a^{2}+u^{2})}{1+u^{2}}}
=2\int_{0}^{\infty} \frac{du}{1-a^{2}+u^{2}}.
$$
这是一个标准积分：
$$
\int_{0}^{\infty} \frac{du}{u^{2}+c^{2}} = \frac{\pi}{2c},\quad c>0.
$$
这里 $c=\sqrt{1-a^{2}}$，所以
$$
I'(a)=2\cdot \frac{\pi}{2\sqrt{1-a^{2}}} = \frac{\pi}{\sqrt{1-a^{2}}}.
$$
又因为 $I(0)=0$，积分得
$$
I(a)=\int_{0}^{a} \frac{\pi}{\sqrt{1-t^{2}}} dt = \pi \arcsin a.
$$
因此
$$
\boxed{I=\pi\arcsin a}.
$$

（2）
设
$$
I(a)=\int_{0}^{\frac{\pi}{2}} \ln\left(\cos^{2}x + a^{2}\sin^{2}x\right)dx,\quad a>0.
$$
对 $a$ 求导：
$$
I'(a)=\int_{0}^{\frac{\pi}{2}} \frac{2a\sin^{2}x}{\cos^{2}x + a^{2}\sin^{2}x} dx.
$$
分子分母同除以 $\cos^{2}x$，令 $t=\tan x$，则 $dx = \frac{dt}{1+t^{2}}$，积分限 $0\to\infty$，
$$
I'(a)=\int_{0}^{\infty} \frac{2a t^{2}}{1+a^{2}t^{2}} \cdot \frac{dt}{1+t^{2}}.
$$
利用部分分式：
$$
\frac{2a t^{2}}{(1+a^{2}t^{2})(1+t^{2})}
= \frac{A}{1+a^{2}t^{2}} + \frac{B}{1+t^{2}}.
$$
通分比较分子：
$$
2a t^{2} = A(1+t^{2}) + B(1+a^{2}t^{2}).
$$
令 $t^{2}=-1$ 得 $ -2a = B(1-a^{2}) \Rightarrow B = \frac{-2a}{1-a^{2}}$。
令 $t^{2}=-\frac{1}{a^{2}}$ 得 $\frac{2a}{a^{2}} = A\left(1-\frac{1}{a^{2}}\right) \Rightarrow \frac{2}{a} = A\frac{a^{2}-1}{a^{2}} \Rightarrow A = \frac{2a}{a^{2}-1}$。
注意 $A = -B$，所以
$$
\frac{2a t^{2}}{(1+a^{2}t^{2})(1+t^{2})}
= \frac{2a}{a^{2}-1}\left( \frac{1}{1+a^{2}t^{2}} - \frac{1}{1+t^{2}} \right).
$$
于是
$$
I'(a)=\frac{2a}{a^{2}-1}\int_{0}^{\infty} \left( \frac{1}{1+a^{2}t^{2}} - \frac{1}{1+t^{2}} \right) dt.
$$
分别积分：
$$
\int_{0}^{\infty} \frac{dt}{1+a^{2}t^{2}} = \frac{1}{a}\cdot\frac{\pi}{2},\quad
\int_{0}^{\infty} \frac{dt}{1+t^{2}} = \frac{\pi}{2}.
$$
所以
$$
I'(a)=\frac{2a}{a^{2}-1}\left( \frac{\pi}{2a} - \frac{\pi}{2} \right)
= \frac{2a}{a^{2}-1}\cdot \frac{\pi}{2}\left( \frac{1}{a} - 1 \right)
= \frac{\pi a}{a^{2}-1}\cdot \frac{1-a}{a}
= \frac{\pi(1-a)}{a^{2}-1}.
$$
注意 $a^{2}-1 = (a-1)(a+1)$，而 $1-a = -(a-1)$，所以
$$
I'(a)=\frac{-\pi(a-1)}{(a-1)(a+1)} = -\frac{\pi}{a+1}.
$$
因此
$$
I(a) = -\pi \ln(a+1) + C.
$$
为确定常数 $C$，取 $a=1$：
$$
I(1)=\int_{0}^{\frac{\pi}{2}} \ln(\cos^{2}x+\sin^{2}x)dx = \int_{0}^{\frac{\pi}{2}} \ln 1\,dx = 0.
$$
代入得
$$
0 = -\pi \ln 2 + C \quad\Rightarrow\quad C = \pi\ln 2.
$$
因此
$$
I(a) = \pi\ln 2 - \pi\ln(a+1) = \pi\ln\frac{2}{a+1}.
$$
故
$$
\boxed{I=\pi\ln\frac{2}{a+1}}.
$$

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