Задача Кузнецов Интегралы 9-2

Условие задачи

Вычислить определенный интеграл:

$\int\limits_{\arccos{\frac{4}{\sqrt{17}}}}^{\pi / 4} \frac{2\operatorname{ctg}{x}+1}{\left(2\sin{x}+\cos{x}\right)^2} dx$

Решение

$\int\limits_{\arccos{\frac{4}{\sqrt{17}}}}^{\frac{\pi}{4}} \frac{2\operatorname{ctg}{x}+1}{\left(2\sin{x}+\cos{x}\right)^2}dx = \int\limits_{\arccos{\frac{4}{\sqrt{17}}}}^{\frac{\pi}{4}} \frac{2\operatorname{ctg}{x} + 1}{(2\operatorname{tg}{x} + 1)^{2}\cdot\cos^{2}{x}}dx = \left|d(\operatorname{tg}{x}) = \frac{dx}{\cos^{2}{x}}\right| =$

$= \int\limits_{\arccos{\frac{4}{\sqrt{17}}}}^{\frac{\pi}{4}}\frac{2\operatorname{ctg}{x} + 1}{(2\operatorname{tg}{x} + 1)^{2}}d(\operatorname{tg}{x})$

Замена:

$t = \operatorname{tg}{x}$

При $x = \arccos{\frac{4}{\sqrt{17}}},~~ t = \frac{1}{4}$

При $x = \frac{\pi}{4},~~ t = 1$

Имеем:

$\int\limits_{\arccos{\frac{4}{\sqrt{17}}}}^{\frac{\pi}{4}}\frac{2\operatorname{ctg}{x} + 1}{(2\operatorname{tg}{x} + 1)^{2}}d(\operatorname{tg}{x}) = \int\limits_{\frac{1}{4}}^{1}\frac{\frac{2}{t} + 1}{(2t + 1)^{2}}dt = \int\limits_{\frac{1}{4}}^{1}\frac{t + 2}{t(2t + 1)^{2}}dt$

$\frac{t + 2}{t(2t + 1)^{2}} = \frac{A}{t} + \frac{B}{2t + 1} + \frac{C}{(2t + 1)^{2}} = \frac{A(2t + 1)^{2} + Bt(2t + 1) + Ct}{t(2t + 1)^{2}}$

$t + 2 = A(2t + 1)^{2} + Bt(2t + 1) + Ct$

$t = 0 \Rightarrow A = 2$

$t = -\frac{1}{2} \Rightarrow -\frac{1}{2}C = -\frac{1}{2} + 2;~~ C = -3$

$t = -1 \Rightarrow A + B - C = 1; ~~ 2 + B + 3 = 1; ~~ B = -4$

$\int\limits_{\frac{1}{4}}^{1}\frac{t + 2}{t(2t + 1)^{2}}dt = \int\limits_{\frac{1}{4}}^{1}\left(\frac{2}{t} - \frac{4}{2t + 1} - \frac{3}{(2t + 1)^{2}}dt\right) = \left. \left(2\ln|t| - 2\ln|2t + 1| + \frac{3}{2}\cdot\frac{1}{2t + 1}\right)\right|_{\frac{1}{4}}^{1}$

$= \left. \left(2\ln\left|\frac{t}{2t + 1}\right| + \frac{3}{2}\cdot\frac{1}{2t + 1}\right)\right|_{\frac{1}{4}}^{1} = 2\ln{\frac{1}{3}} + \frac{1}{2} - 2\ln{\frac{1}{6}} - 1 = 2\ln{2} - \frac{1}{2}$