Одређени интеграл 2

Задаци

Текст задатака објашњених у видео лекцији:

Пр.3)   Решити   $\int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\left( {\frac{1}{{{{\cos }^2}x}} - \sin x} \right)dx}$

Пр.4)   $\int\limits_{ - 2}^{ - 1} {\frac{{dx}}{{{{\left( {11 + 5x} \right)}^3}}}}$

Пр.5)   $\int\limits_2^{ - 13} {\frac{{dx}}{{\sqrt[5]{{3 - {x^4}}}}}}$

Пр.3)   $\int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\left( {\frac{1}{{{{\cos }^2}x}} - \sin x} \right)dx} = \int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\frac{1}{{{{\cos }^2}x}}dx} - \int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\sin xdx} = \left. {tgx} \right|_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} + \left. {\cos x} \right|_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} =$

$= \left( {tgx + \left. {\cos x} \right|} \right)_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} = tg\frac{\pi }{4} + \cos \frac{\pi }{4} - \left( {tg\frac{\pi }{4} + \cos \frac{\pi }{4}} \right) =$

$= 1 + \frac{{\sqrt 2 }}{2} - \left( { - 1 + \frac{{\sqrt 2 }}{2}} \right) = 2$

Пр.4)   $\int\limits_{ - 2}^{ - 1} {\frac{{dx}}{{{{\left( {11 + 5x} \right)}^3}}}}$

$$\begin{array}{cccccccccccccccccccc} \begin{aligned}&11 + 5x = t \\ &5dx = dt \\ &dx = \frac{{dt}}{5}\end{aligned} &\begin{aligned}&x = - 2 \\ &x = - 1\end{aligned} &\begin{aligned}&t = 11 + 5\left( { - 2} \right) = 1 \\ &t = 11 + 5\left( { - 1} \right) = 6\end{aligned} \end{array}$$

$= \int\limits_1^6 {\frac{{\frac{{dt}}{5}}}{{{t^3}}}} = \frac{1}{5}\int\limits_1^6 {{t^{ - 3}}dt = } \frac{1}{5} \cdot \left. {\frac{{{t^{ - 2}}}}{{ - 2}}} \right|_1^6 = \frac{{ - 1}}{{10}} \cdot \left. {\frac{1}{{{t^2}}}} \right|_1^6 = - \frac{1}{{10}} \cdot \left( {\frac{1}{{{6^2}}} - \frac{1}{{{1^2}}}} \right) = - \frac{1}{{10}} \cdot \left( { - \frac{{35}}{{36}}} \right) = \frac{7}{{72}}$

Пр.5)   $\int\limits_2^{ - 13} {\frac{{dx}}{{\sqrt[5]{{3 - {x^4}}}}}}$

$$\begin{array}{cccccccccccccccccccc} \begin{aligned}&3 - x = t \\ &- dx = dt \\ &dx = - dt\end{aligned} &\begin{aligned}&x = 2 \\ &x = - 13\end{aligned} &\begin{aligned}&t = 1 \\ &t = 16\end{aligned} \end{array}$$

$= \int\limits_1^{16} {\frac{{ - dt}}{{{{\sqrt[5]{t}}^4}}}} = - \int\limits_1^{16} {{t^{ - \frac{4}{3}}}} dt = - \left. {\frac{{{t^{\frac{1}{5}}}}}{{\frac{1}{5}}}} \right|_1^{16} = - 5\left. {\sqrt[5]{t}} \right|_1^{16} = - 5\sqrt[5]{{16}} - \left( { - 5\sqrt[5]{1}} \right) = - 5\left( {\sqrt[5]{{16}} - 1} \right)$