2020-09-10 积分题

zedo2020/9/10大约 2 分钟约 497 字

题目

计算积分 $\int_{0}^{\frac{π}{2}} \frac{sin ^{3} x cos x}{1 + sin x cos x} d x$ .

此题解法也是笔者在知乎 ^[1] 的回答。

解法

利用区间再现，凑微分即可：

I x \mapsto \frac{π}{2} - x \int_{0}^{\frac{π}{2}} \frac{cos x sin x cos x}{1 + sin x cos x} d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{( sin x + cos x ) - \frac{1}{2} ( sin x - cos x ) ^{2} + \frac{1}{2}}{\frac{3}{2} - \frac{1}{2} ( sin x - cos x ) ^{2}} d x = \frac{2}{2} \int_{0}^{\frac{π}{2}} \frac{1 - ( sin x - cos x ) ^{2}}{3 - ( sin x - cos x ) ^{2}} d (sin x - cos x) = \frac{2}{2} \int_{- 1}^{1} \frac{1 - t ^{2}}{3 - t ^{2}} d t = 2 \int_{0}^{1} \frac{1 - t ^{2}}{3 - t ^{2}} d t t = s i n u 2 \int_{0}^{\frac{π}{2}} \frac{cos ^{2} u}{3 - sin ^{2} u} d u = 2 \int_{0}^{\frac{π}{2}} (1 - \frac{2}{3 - sin ^{2} u}) d u = \frac{π}{2} - 22 \int_{0}^{\frac{π}{2}} \frac{d ( tan u )}{3 ( 1 + tan ^{2} u ) - tan ^{2} u} = \frac{π}{2} - 22 \int_{0}^{\frac{π}{2}} \frac{d ( tan u )}{3 + 2 tan ^{2} u} = \frac{π}{2} - \frac{2 2}{2 3} [arctan \frac{2 tan u}{3}]_{0}^{\frac{π}{2}} = \frac{π}{2} - \frac{π}{3}

利用 Cauchy-Schlömilch 变换 ^[2]：

I = \int_{0}^{\frac{π}{2}} \frac{tan x tan x}{sec ^{2} x + tan x} d x t = t a n x \int_{0}^{\infty} \frac{t ^{3}}{1 + t ^{4} + t ^{2}} \cdot \frac{2 t}{1 + t ^{4}} d t = \int_{- \infty}^{+ \infty} \frac{1}{\frac{1}{t ^{2}} + t ^{2} + 1} \frac{1}{\frac{1}{t ^{2}} + t ^{2}} d t = \int_{- \infty}^{+ \infty} \frac{1}{( t - \frac{1}{t} ) ^{2} + 3} \frac{1}{( t - \frac{1}{t} ) ^{2} + 2} d t Cauchy-Schl \overset{o}{¨} milch \int_{- \infty}^{+ \infty} \frac{1}{t ^{2} + 3} \frac{1}{t ^{2} + 2} d t = 2 \int_{0}^{+ \infty} (\frac{1}{t ^{2} + 2} - \frac{1}{t ^{2} + 3}) d t = \frac{π}{2} - \frac{π}{3}