Упр.1011 ГДЗ Алимов 10-11 класс (Алгебра)

$$\int_{-\pi}^{\pi}\sin^2 x\,dx=\int_{-\pi}^{\pi}\frac{1-\cos 2x}{2}\,dx$$

$$=\left(\frac{x}{2}-\frac{\sin 2x}{4}\right)\Bigg|_{-\pi}^{\pi}=\frac{\pi}{2}-0-\left(-\frac{\pi}{2}-0\right)=\pi$$

$$\int_{0}^{\pi/2}\sin x\cos x\,dx=\int_{0}^{\pi/2}\frac{1}{2}\sin 2x\,dx$$

$$=-\frac{1}{4}\cos 2x\Bigg|_{0}^{\pi/2}=-\frac{\cos\pi}{4}+\frac{\cos 0}{4}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$

$$\int_{0}^{\pi/4}(\cos^2 x-\sin^2 x)\,dx=\int_{0}^{\pi/4}\cos 2x\,dx$$

$$=\frac{1}{2}\sin 2x\Bigg|_{0}^{\pi/4}=\frac{1}{2}\sin\frac{\pi}{2}-\frac{1}{2}\sin 0=\frac{1}{2}$$

$$\int_{0}^{\pi}(\sin^4 x+\cos^4 x)\,dx$$

$$=\int_{0}^{\pi}\left((\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x\right)\,dx$$

$$=\int_{0}^{\pi}\left(1-\frac{1}{2}\sin^2 2x\right)\,dx=\int_{0}^{\pi}\left(1-\frac{1-\cos 4x}{4}\right)\,dx$$

$$=\int_{0}^{\pi}\left(\frac{3}{4}+\frac{\cos 4x}{4}\right)\,dx=\left(\frac{3x}{4}+\frac{\sin 4x}{16}\right)\Bigg|_{0}^{\pi}=\frac{3\pi}{4}$$

$$\int_{0}^{3}x^2\sqrt{x+1}\,dx=\int_{0}^{3}(x+1-1)^2\sqrt{x+1}\,dx$$

$$=\int_{0}^{3}\left((x+1)^{5/2}-2(x+1)^{3/2}+(x+1)^{1/2}\right)\,dx$$

$$=\left(\frac{2}{7}(x+1)^{7/2}-\frac{4}{5}(x+1)^{5/2}+\frac{2}{3}(x+1)^{3/2}\right)\Bigg|_{0}^{3}$$

$$=\frac{2}{7}\cdot 4^{7/2}-\frac{4}{5}\cdot 4^{5/2}+\frac{2}{3}\cdot 4^{3/2}-\left(\frac{2}{7}-\frac{4}{5}+\frac{2}{3}\right)=\frac{116}{15}$$

$$\int_{3}^{4}\frac{x^2-4x+5}{x-2}\,dx=\int_{3}^{4}\frac{(x-2)^2+1}{x-2}\,dx$$

$$=\int_{3}^{4}\left(x-2+\frac{1}{x-2}\right)\,dx$$

$$=\left(\frac{x^2}{2}-2x+\ln(x-2)\right)\Bigg|_{3}^{4}$$

$$=\left(8-8+\ln 2\right)-\left(\frac{9}{2}-6+\ln 1\right)=\frac{3}{2}+\ln 2$$