ГДЗ по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 73

\[\boxed{\mathbf{73}.}\]

\[\textbf{а)}\ \int_{0}^{\frac{\pi}{2}}{\cos{2x}\text{dx}};\]

\[u = 2x;\ \ x = \frac{u}{2};\]

\[dx = d\left( \frac{u}{2} \right) = \frac{1}{2}du;\]

\[x = 0 \rightarrow u = 0;\]

\[x = \frac{\pi}{2} \rightarrow u = 2 \cdot \frac{\pi}{2} = \pi;\]

\[\int_{0}^{\pi}{\cos u\frac{1}{2}\text{du}} = \frac{1}{2}\int_{0}^{\pi}{\cos u\text{du}} =\]

\[= \frac{1}{2}\left( F(\pi) - F(0) \right) =\]

\[= \frac{1}{2}\left( \sin\pi - \sin 0 \right) =\]

\[= \frac{1}{2}(0 - 0) = 0.\]

\[\textbf{б)}\ \int_{0}^{\pi}{\sin\frac{x}{3}\text{dx}};\]

\[u = \frac{x}{3};\ \ x = 3u;\]

\[dx = d \cdot (3u) = 3du;\]

\[x = 0 \rightarrow u = \frac{0}{3} = 0;\]

\[x = \pi \rightarrow u = \frac{\pi}{3};\]

\[\int_{0}^{\frac{\pi}{3}}{\sin u3du} = 3\int_{0}^{\frac{\pi}{3}}{\sin u\text{du}} =\]

\[= 3\left( F\left( \frac{\pi}{3} \right) - F(0) \right) =\]

\[= 3\left( - \cos\frac{\pi}{3} + \cos 0 \right) =\]

\[= 3 \cdot \left( - \frac{1}{2} + 1 \right) = \frac{3}{2} = 1,5.\]

\[\textbf{в)}\ \int_{- 1}^{1}{\sqrt{1 - x^{2}}\text{dx}};\]

\[x = \sin u;u = \arcsin x;\]

\[dx = d\left( \sin u \right) = \cos udu;\]

\[x = - 1 \rightarrow u = \arcsin( - 1) =\]

\[= - \frac{\pi}{2};\]

\[x = 1 \rightarrow u = \arcsin 1 = \frac{\pi}{2};\]

\[\int_{- 1}^{1}{\sqrt{1 - x^{2}}\text{dx}} = \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}}{\text{co}s^{2}\text{u\ du}} =\]

\[= \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}}{\frac{1 + \cos{2u}}{2}\text{du}} =\]

\[\textbf{г)}\ \int_{0}^{2}{\sqrt{4 - x^{2}}\text{dx}};\]

\[x = 2\sin u;\]

\[u = \arcsin\frac{x}{2};\]

\[dx = 2\cos udu;\]

\[x = 0 \rightarrow u = \arcsin 0 = 0;\]

\[x = 2 \rightarrow u = \arcsin\frac{2}{2} =\]

\[= \arcsin 1 = \frac{\pi}{2};\]

\[\int_{0}^{2}{\sqrt{4 - x^{2}}\text{dx}} =\]

\[= \int_{0}^{\frac{\pi}{2}}{\sqrt{4 - 4sin^{2}u} \cdot 2\cos u\text{du}} =\]

\[= \int_{0}^{\frac{\pi}{2}}{2\sqrt{1 - sin^{2}u} \cdot 2\cos u\text{du}} =\]

\[= \int_{0}^{\frac{\pi}{2}}{4\cos^{2}u\text{du}} =\]

\[= \int_{0}^{\frac{\pi}{2}}{4 \cdot \frac{1 + \cos{2u}}{2}\text{du}} =\]

\[= \int_{0}^{\frac{\pi}{2}}{2 \cdot \left( 1 + \cos{2u} \right)\text{du}} =\]

\[= 2 \cdot \int_{0}^{\frac{\pi}{2}}{\left( 1 + \cos{2u} \right)\text{du}} =\]

\[= 2 \cdot \left( \frac{\pi}{2} + 0 \right) = \pi.\]

\[\textbf{д)}\ \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\text{dx}}{4 + x^{2}};\]

\[x = 2u;\ \]

\[dx = 2du;\]

\[\int_{- \frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\text{dx}}{4 + x^{2}} = \int_{- \frac{\pi}{4}}^{\frac{\pi}{4}}\frac{2du}{4 + 4u^{2}} =\]

\[= \frac{1}{2}\int_{- \frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\text{du}}{1 + u^{2}} = \left. \ \frac{1}{2}\text{artg\ u} \right|_{- \frac{\pi}{4}}^{\frac{\pi}{4}} =\]

\[= \frac{1}{2}\text{artg\ }\frac{\pi}{4} - \frac{1}{2}\text{artg}\left( - \frac{\pi}{4} \right) =\]

\[= artg\ \frac{\pi}{4}.\]

\[\textbf{е)}\ \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\text{dx}}{\sqrt{9 - x^{2}}};\]

\[x = 3\sin u;\ \]

\[u = \arcsin\frac{x}{3};\]

\[dx = 3\cos udu;\]

\[x = - \frac{\pi}{2} \rightarrow u = \arcsin\left( - \frac{\pi}{6} \right) =\]

\[= - \arcsin\frac{\pi}{6};\]

\[x = \frac{\pi}{2} \rightarrow u = \arcsin\frac{\pi}{6};\]

\[\int_{- \frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\text{dx}}{\sqrt{9 - x^{2}}} =\]

\[= \int_{- \arcsin\frac{\pi}{6}}^{\arcsin\frac{\pi}{6}}\frac{3\cos u\text{du}}{\sqrt{9 - 9sin^{2}u}} =\]

\[= \int_{- \arcsin\frac{\pi}{6}}^{\arcsin\frac{\pi}{6}}\frac{3\cos u\text{du}}{3\sqrt{1 - sin^{2}u}} =\]

\[= \int_{- \arcsin\frac{\pi}{6}}^{\arcsin\frac{\pi}{6}}\frac{\cos u\text{du}}{\cos u} =\]

\[= \int_{- \arcsin\frac{\pi}{6}}^{\arcsin\frac{\pi}{6}}{1du} =\]

\[= \arcsin\frac{\pi}{6} - \left( - \arcsin\frac{\pi}{6} \right) =\]

\[= 2\arcsin\frac{\pi}{6}.\]