Упр.21.10 ГДЗ Мордковича 11 класс профильный уровень (Алгебра)
- 21.10. Вычислите интеграл:
а) $$\int_{0}^{\frac{3\pi}{4}} 6\sin\left(\frac{x}{3}\right)\cos\left(\frac{x}{3}\right)\,dx$$;
в) $$\int_{0}^{-\frac{\pi}{8}}\left(\sin^2(2x)-\cos^2(2x)\right)\,dx$$;
б) $$\int_{-\pi}^{0}\left(2\sin\left(\frac{x}{4}\right)-1\right)\,dx$$;
г) $$\int_{0}^{-\frac{3\pi}{4}}\left(1-2\cos^2\left(\frac{x}{3}\right)\right)\,dx$$.
а) $$\int\limits_{0}^{\frac{3\pi}{4}} 6\sin\frac{x}{3}\cos\frac{x}{3}\,dx=\int\limits_{0}^{\frac{3\pi}{4}} 3\sin\frac{2x}{3}\,dx$$
$$=3\cdot\left(-\frac{3}{2}\cos\frac{2x}{3}\right)\Bigg|_{0}^{\frac{3\pi}{4}}=-\frac{9}{2}\cos\frac{2x}{3}\Bigg|_{0}^{\frac{3\pi}{4}}$$
$$=-\frac{9}{2}\cos\frac{\pi}{2}+\frac{9}{2}\cos 0=\frac{9}{2}=4{,}5.$$
б) $$\int\limits_{-\pi}^{0}\left(2\sin^2\frac{x}{4}-1\right)\,dx=\int\limits_{-\pi}^{0}\left(2\cdot\frac{1-\cos\frac{x}{2}}{2}-1\right)\,dx$$
$$=\int\limits_{-\pi}^{0}\left(-\cos\frac{x}{2}\right)\,dx=-2\sin\frac{x}{2}\Bigg|_{-\pi}^{0}$$
$$=-2\sin 0+2\sin\frac{\pi}{2}=-2.$$
в) $$\int\limits_{0}^{-\frac{\pi}{8}}\left(\sin^2 2x-\cos^2 2x\right)\,dx=\int\limits_{0}^{-\frac{\pi}{8}}(-\cos 4x)\,dx$$
$$=-\frac14\sin 4x\Bigg|_{0}^{-\frac{\pi}{8}}=-\frac14\sin\left(-\frac{\pi}{2}\right)+\frac14\sin 0=\frac14=0{,}25.$$
г) $$\int\limits_{0}^{-\frac{3\pi}{4}}\left(1-2\cos^2\frac{x}{3}\right)\,dx=\int\limits_{0}^{-\frac{3\pi}{4}}\left(1-2\cdot\frac{1+\cos\frac{2x}{3}}{2}\right)\,dx$$
$$=\int\limits_{0}^{-\frac{3\pi}{4}}\left(-\cos\frac{2x}{3}\right)\,dx=-\frac32\sin\frac{2x}{3}\Bigg|_{0}^{-\frac{3\pi}{4}}$$
$$=-\frac32\sin\left(-\frac{\pi}{2}\right)+\frac32\sin 0=\frac32=1{,}5.$$
Ответ
а) $$4{,}5$$; б) $$-2$$; в) $$0{,}25$$; г) $$1{,}5$$.
