Упр.11.20 ГДЗ Мерзляк 11 класс Базовый уровень (Алгебра)

1) ?(5?/4; 15?/4)ctg^2(x/5)dx; 3) ?(-?/2; -?/4)sin(7x)cos(3x)dx;
2) ?(-?/2; ?/2)cos^2(2x)dx; 4) ?(-2; -1)(x^2-e^x)/(x^2 e^x)dx.

1. $$\int\limits_{\frac{5\pi}{4}}^{\frac{15\pi}{4}} \ctg^2\frac{x}{5}\,dx
   =\int\limits_{\frac{5\pi}{4}}^{\frac{15\pi}{4}}\left(\frac{1}{\sin^2\frac{x}{5}}-1\right)\,dx.$$
   Тогда
   $$\int \ctg^2\frac{x}{5}\,dx=-5\ctg\frac{x}{5}-x.$$
   Следовательно,
   $$
   \int\limits_{\frac{5\pi}{4}}^{\frac{15\pi}{4}} \ctg^2\frac{x}{5}\,dx
   =\left(-5\ctg\frac{x}{5}-x\right)\Bigg|_{\frac{5\pi}{4}}^{\frac{15\pi}{4}}.
   $$
   Подставим пределы:
   $$
   \left(-5\ctg\frac{3\pi}{4}-\frac{15\pi}{4}\right)-\left(-5\ctg\frac{\pi}{4}-\frac{5\pi}{4}\right)
   =\left(5-\frac{15\pi}{4}\right)-\left(-5-\frac{5\pi}{4}\right)
   =10-\frac{10\pi}{4}
   =\frac{20-5\pi}{2}.
   $$

2. $$\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 2x\,dx
   =\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\cos 4x}{2}\,dx.$$
   Тогда
   $$
   \int \frac{1+\cos 4x}{2}\,dx=\frac{1}{2}\left(x+\frac{1}{4}\sin 4x\right).
   $$
   Значит,
   $$
   \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 2x\,dx
   =\frac{1}{2}\left(x+\frac{1}{4}\sin 4x\right)\Bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
   =\frac{1}{2}\left(\frac{\pi}{2}+\frac{1}{4}\sin 2\pi\right)
   -\frac{1}{2}\left(-\frac{\pi}{2}+\frac{1}{4}\sin(-2\pi)\right)
   =\frac{\pi}{2}.
   $$

3. $$\int\limits_{-\frac{\pi}{2}}^{-\frac{\pi}{4}} \sin 7x\cos 3x\,dx
   =\int\limits_{-\frac{\pi}{2}}^{-\frac{\pi}{4}} \frac{1}{2}\bigl(\sin 10x+\sin 4x\bigr)\,dx.$$
   Тогда
   $$
   \int \frac{1}{2}\bigl(\sin 10x+\sin 4x\bigr)\,dx
   =\frac{1}{2}\left(-\frac{1}{10}\cos 10x-\frac{1}{4}\cos 4x\right).
   $$
   Следовательно,
   $$
   \int\limits_{-\frac{\pi}{2}}^{-\frac{\pi}{4}} \sin 7x\cos 3x\,dx
   =\left(-\frac{1}{20}\cos 10x-\frac{1}{8}\cos 4x\right)\Bigg|_{-\frac{\pi}{2}}^{-\frac{\pi}{4}}.
   $$
   Подставим пределы:
   $$
   \left(-\frac{1}{20}\cos\frac{5\pi}{2}-\frac{1}{8}\cos\pi\right)
   -\left(-\frac{1}{20}\cos(-5\pi)-\frac{1}{8}\cos(-2\pi)\right)
   =\left(0+\frac{1}{8}\right)-\left(\frac{1}{20}-\frac{1}{8}\right)
   =\frac{1}{5}.
   $$

4. $$\int\limits_{-2}^{-1}\frac{x^2-e^x}{x^2e^x}\,dx
   =\int\limits_{-2}^{-1}\left(\frac{1}{e^x}-\frac{1}{x^2}\right)\,dx
   =\int\limits_{-2}^{-1}\left(e^{-x}-x^{-2}\right)\,dx.$$
   Тогда
   $$
   \int \left(e^{-x}-x^{-2}\right)\,dx=-e^{-x}+\frac{1}{x}.
   $$
   Значит,
   $$
   \int\limits_{-2}^{-1}\frac{x^2-e^x}{x^2e^x}\,dx
   =\left(-e^{-x}+\frac{1}{x}\right)\Bigg|_{-2}^{-1}
   =\left(-e+\left(-1\right)\right)-\left(-e^2-\frac{1}{2}\right)
   =e^2-e-\frac{1}{2}.
   $$

Ответ

1) $$\frac{20-5\pi}{2}$$; 2) $$\frac{\pi}{2}$$; 3) $$\frac{1}{5}$$; 4) $$e^2-e-\frac{1}{2}$$.