Упр.1037 ГДЗ Алимов 10-11 класс (Алгебра)
- $$\int_{0}^{\frac{\pi}{2}} \frac{1}{2}\cos\left(x+\frac{\pi}{4}\right)\,dx$$
- $$\int_{0}^{\frac{\pi}{3}} \frac{1}{3}\sin\left(x-\frac{\pi}{3}\right)\,dx$$
- $$\int_{1}^{3} \sin(3x-6)\,dx$$
- $$\int_{0}^{3} 8\cos(4x-12)\,dx$$
1) $$\int\limits_{0}^{\pi/4}\frac12\cos\left(x+\frac{\pi}{4}\right)\,dx=\frac12\sin\left(x+\frac{\pi}{4}\right)\Bigg|_{0}^{\pi/4}$$
$$=\frac12\left(\sin\frac{\pi}{2}-\sin\frac{\pi}{4}\right)=\frac12\left(1-\frac{\sqrt2}{2}\right)=\frac{2-\sqrt2}{4}.$$
2) $$\int\limits_{0}^{\pi/3}\frac13\sin\left(x-\frac{\pi}{3}\right)\,dx=-\frac13\cos\left(x-\frac{\pi}{3}\right)\Bigg|_{0}^{\pi/3}$$
$$=-\frac13\left(\cos 0-\cos\left(-\frac{\pi}{3}\right)\right)=-\frac13\left(1-\frac12\right)=-\frac16.$$
3) $$\int\limits_{1}^{3}3\sin(3x-6)\,dx=-\cos(3x-6)\Bigg|_{1}^{3}$$
$$=-\cos 3+\cos(-3)=0.$$
4) $$\int\limits_{0}^{3}8\cos(4x-12)\,dx=2\sin(4x-12)\Bigg|_{0}^{3}$$
$$=2\bigl(\sin 0-\sin(-12)\bigr)=2\sin 12.$$
Ответ: 1) $$\frac{2-\sqrt2}{4}$$; 2) $$-\frac16$$; 3) $$0$$; 4) $$2\sin 12$$.









