📝 题目
例 10 求不定积分 $\displaystyle{I = \int \frac{{\cos }^{3}x}{\cos x + \sin x}\mathrm{\;d}x,J = \int \frac{{\sin }^{3}x}{\cos x + \sin x}\mathrm{\;d}x}$ .
💡 答案与解析
解 $I + J = \int \left( {1 - \frac{1}{2}\sin {2x}}\right) \mathrm{d}x = x + \frac{1}{4}\cos {2x} + C$ ,
$$ I - J = \int \frac{{\cos }^{3}x - {\sin }^{3}x}{\cos x + \sin x}\mathrm{\;d}x $$
$$ = \int \frac{\left( {\cos x - \sin x}\right) \left( {1 + \frac{1}{2}\sin {2x}}\right) }{\cos x + \sin x}\mathrm{\;d}x $$
$$ = \int \frac{\left( {{\cos }^{2}x - {\sin }^{2}x}\right) \left( {1 + \frac{1}{2}\sin {2x}}\right) }{{\left( \cos x + \sin x\right) }^{2}}\mathrm{\;d}x $$
$$ = \int \frac{\left( {1 + \frac{1}{2}\sin {2x}}\right) \cos {2x}}{1 + \sin {2x}}\mathrm{\;d}x $$
$$ = \frac{1}{4}\sin {2x} + \frac{1}{4}\ln \left( {\sin {2x} + 1}\right) + C, $$
由此求得
$$ I = \frac{1}{2}x + \frac{1}{8}\cos {2x} + \frac{1}{8}\sin {2x} + \frac{1}{8}\ln \left( {2\sin {2x} + 2}\right) + C, $$
$$ J = \frac{1}{2}x + \frac{1}{8}\cos {2x} - \frac{1}{8}\sin {2x} - \frac{1}{8}\ln \left( {2\sin {2x} + 2}\right) + C. $$
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