📝 题目
例 8 求 $\displaystyle{\int \frac{\mathrm{d}x}{\cos x}}$ .
💡 答案与解析
解 $\displaystyle{\int \frac{\mathrm{d}x}{\cos x} = \int \frac{\cos x\mathrm{\;d}x}{{\cos }^{2}x}}$
$$ = \int \frac{\mathrm{d}\left( {\sin x}\right) }{1 - {\sin }^{2}x} $$
$$ = \frac{1}{2}\ln \left| \frac{1 + \sin x}{1 - \sin x}\right| + C $$
$$ = \frac{1}{2}\ln {\left| \frac{1 + \sin x}{\cos x}\right| }^{2} + C $$
$$ = \ln \left| {\frac{1}{\cos x} + \frac{\sin x}{\cos x}}\right| + C $$
$$ = \ln \left| {\sec x + \tan x}\right| + C\text{ , } $$
在计算过程中, 我们用到以下事实:
$$ \int \frac{\mathrm{d}u}{{u}^{2} - 1} = \frac{1}{2}\ln \left| \frac{u - 1}{u + 1}\right| + C. $$