Решебник по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 23

\[\boxed{\mathbf{23}.}\]

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

\[x = \sin t;\ \]

\[dx = \cos tdt;\]

\[\int_{}^{}{\sqrt{1 - \left( \sin t \right)^{2}}\cos t\text{dt}} =\]

\[= \int_{}^{}{\cos^{2}t\text{dt}} =\]

\[= \int_{}^{}{\frac{1 + \cos{2t}}{2}\text{dt}} =\]

\[= \frac{1}{2}\int_{}^{}\text{dt} + \frac{1}{2}\int_{}^{}{\cos{2t}\text{dt}} =\]

\[= \frac{1}{2}t + \frac{1}{4}\sin{2t} + C;\]

\[t = \arcsin x:\]

\[= \frac{1}{2}\arcsin x + \frac{1}{2}x\sqrt{1 - x^{2}} + C.\]

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

\[x = 2\sin t;\]

\[dx = 2\cos tdt;\]

\[\int_{}^{}{\sqrt{4 - \left( 2\sin t \right)^{2}}{2cos}t\text{dt}} =\]

\[= 4\int_{}^{}{\cos^{2}t\text{dt}} =\]

\[= 4\int_{}^{}{\frac{1 + \cos{2t}}{2}\text{dt}} =\]

\[= 2\int_{}^{}\text{dt} + 2\int_{}^{}{\cos{2t}\text{dt}} =\]

\[= 2t + \frac{1}{2}\sin{2t} + C;\]

\[t = \arcsin\frac{x}{2}:\]

\[= 2\arcsin\frac{x}{2} + \frac{x\sqrt{4 - x^{2}}}{4} + C.\]

\[\textbf{в)}\ \int_{}^{}{\sqrt{1 - 4x^{2}}\text{dx}}\]

\[x = \frac{\sin t}{2};\]

\[dx = \frac{\cos t}{2}dt;\]

\[\int_{}^{}{\sqrt{1 - 4x^{2}}\text{dt}} = \int_{}^{}{\frac{\cos^{2}t}{2}\text{dt}} =\]

\[= \frac{1}{2}\int_{}^{}{\cos^{2}t\text{dt}} =\]

\[= \frac{1}{2}\int_{}^{}{\frac{1 + \cos{2t}}{2}\text{dt}} =\]

\[= \frac{1}{4}\int_{}^{}{\left( 1 + \cos{2t} \right)\text{dt}} =\]

\[= \frac{1}{4}\int_{}^{}\text{dt} + \frac{1}{4}\int_{}^{}{\left( \cos{2t} \right)\text{dt}} =\]

\[= \frac{1}{4}t + \frac{1}{8}\sin{2t} + C;\]

\[t = \arcsin{2x}:\]

\[= \frac{1}{4}\arcsin{2x} + \frac{x\sqrt{1 - 4x^{2}}}{2} + C.\]

\[\textbf{г)}\ \int_{}^{}{\sqrt{1 - 9x^{2}}\text{dx}}\]

\[x = \frac{\sin t}{3};\]

\[dx = \frac{\cos t}{3}dt;\]

\[\int_{}^{}{\frac{\cos^{2}t}{2}\text{dt}} = \frac{1}{3}\int_{}^{}{\cos^{2}t\text{dt}} =\]

\[= \frac{1}{3}\int_{}^{}{\frac{1 + \cos{2t}}{2}\text{dt}} =\]

\[= \frac{1}{6}\int_{}^{}{\left( 1 + \cos{2t} \right)\text{dt}} =\]

\[= \frac{1}{6}\int_{}^{}\text{dt} + \frac{1}{6}\int_{}^{}{\left( \cos{2t} \right)\text{dt}} =\]

\[= \frac{1}{6}t + \frac{1}{12}\sin{2t} + C;\]

\[t = \arcsin{2x}:\]

\[= \frac{1}{6}\arcsin{3x} + \frac{x\sqrt{1 - 9x^{2}}}{2} + C.\]