Упр.375 ГДЗ Колягин Ткачёва 11 класс (Алгебра)

2) интеграл (пи/4;пи) cos(3x-пи/4) dx;
3) интеграл (0;пи/4) cos2 4x dx;
4) интеграл (0;пи/3) sin2(x-пи/3) dx.

1. $$\int\limits_{0}^{\pi/2}\sin 2x\,dx=-\frac12\cos 2x\Big|_{0}^{\pi/2}$$
   $$=-\frac12\cos\pi+\frac12\cos 0=\frac12+\frac12=1.$$

2. $$\int\limits_{\pi/4}^{\pi}\cos\left(3x-\frac{\pi}{4}\right)\,dx=\frac13\sin\left(3x-\frac{\pi}{4}\right)\Big|_{\pi/4}^{\pi}$$
   $$=\frac13\sin\frac{11\pi}{4}-\frac13\sin\frac{\pi}{2}=\frac13\cdot\frac{\sqrt2}{2}-\frac13=\frac{\sqrt2}{6}-\frac13.$$

3. $$\int\limits_{0}^{\pi/4}\cos^2 4x\,dx=\int\limits_{0}^{\pi/4}\frac{1+\cos 8x}{2}\,dx$$
   $$=\int\limits_{0}^{\pi/4}\left(\frac12+\frac12\cos 8x\right)\,dx$$
   $$=\left(\frac12x+\frac1{16}\sin 8x\right)\Big|_{0}^{\pi/4}$$
   $$=\frac12\cdot\frac{\pi}{4}+\frac1{16}\sin 2\pi-\left(\frac12\cdot 0+\frac1{16}\sin 0\right)=\frac{\pi}{8}.$$

4. $$\int\limits_{0}^{\pi/3}\sin^2\left(x-\frac{\pi}{3}\right)\,dx=\int\limits_{0}^{\pi/3}\frac{1-\cos\left(2x-\frac{2\pi}{3}\right)}{2}\,dx$$
   $$=\int\limits_{0}^{\pi/3}\left(\frac12-\frac12\cos\left(2x-\frac{2\pi}{3}\right)\right)\,dx$$
   $$=\left(\frac12x-\frac14\sin\left(2x-\frac{2\pi}{3}\right)\right)\Big|_{0}^{\pi/3}$$
   $$=\frac{\pi}{6}-\frac14\sin 0-\left(0-\frac14\sin\left(-\frac{2\pi}{3}\right)\right)$$
   $$=\frac{\pi}{6}-\frac14\cdot\frac{\sqrt3}{2}=\frac{\pi}{6}-\frac{\sqrt3}{8}=\frac{4\pi-3\sqrt3}{24}.$$

Ответ

1) $$1$$; 2) $$\frac{\sqrt2}{6}-\frac13$$; 3) $$\frac{\pi}{8}$$; 4) $$\frac{4\pi-3\sqrt3}{24}$$.