Решебник по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 16

\[\boxed{\mathbf{16}.}\]

\[\textbf{а)}\ \int_{}^{}\frac{\text{dx}}{1 + \cos{2x}} = \int_{}^{}\frac{\text{dx}}{2cos^{2}x} =\]

\[= \frac{1}{2}\int_{}^{}\frac{\text{dx}}{\text{co}s^{2}x} = \frac{1}{2}tgx + C.\]

\[\textbf{б)}\ \int_{}^{}\frac{\text{dx}}{1 - \cos{2x}} = \int_{}^{}\frac{\text{dx}}{2sin^{2}x} =\]

\[= \frac{1}{2}\int_{}^{}\frac{\text{dx}}{\text{si}n^{2}x} = - \frac{1}{2}ctgx + C.\]

\[\textbf{в)}\ \int_{}^{}{\left( \text{co}s^{2}x - sin^{2}x \right)\text{dx}} =\]

\[= \int_{}^{}{\cos{2x}\text{dx}} = \frac{1}{2}\sin{2x} + C.\]

\[\textbf{г)}\ \int_{}^{}{\sin x\cos x\text{dx}} =\]

\[= \int_{}^{}{\frac{1}{2}\sin{2x}\text{dx}} =\]

\[= \frac{1}{2}\int_{}^{}{\sin{2x}\text{dx}} =\]

\[= \frac{1}{2} \cdot \left( - \frac{1}{2} \right)\cos{2x} + C =\]

\[= - \frac{1}{4}\cos{2x} + C.\]