Упр.1037 ГДЗ Алимов 10-11 класс (Алгебра)

1. $$\int_{0}^{\frac{\pi}{2}} \frac{1}{2}\cos\left(x+\frac{\pi}{4}\right)\,dx$$
2. $$\int_{0}^{\frac{\pi}{3}} \frac{1}{3}\sin\left(x-\frac{\pi}{3}\right)\,dx$$
3. $$\int_{1}^{3} \sin(3x-6)\,dx$$
4. $$\int_{0}^{3} 8\cos(4x-12)\,dx$$

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) $$2\sin 12$$.