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

Задача

2) интеграл (0;пи/2) sinxcosxdx;
3) интеграл (0;пи/4) (cos2x-sin2x)dx;
4) интеграл (0;пи) (sin4x+cos4x)dx;
5) интеграл (0;3) x2 корень (x+1)dx;
6) интеграл (3;4) (x2-4x+5))dx/(x-2).

Подробный ответ

$$\int\limits_{-\pi}^{\pi}\sin^2 x\,dx=\int\limits_{-\pi}^{\pi}\frac{1-\cos 2x}{2}\,dx$$
$$=\left(\frac{x}{2}-\frac{\sin 2x}{4}\right)\Bigg|_{-\pi}^{\pi}=\frac{\pi}{2}-0-\left(-\frac{\pi}{2}-0\right)=\pi$$
$$\int\limits_{0}^{\pi/2}\sin x\cos x\,dx=\frac12\int\limits_{0}^{\pi/2}\sin 2x\,dx$$
$$=\left(-\frac{\cos 2x}{4}\right)\Bigg|_{0}^{\pi/2}=-\frac{\cos\pi}{4}+\frac{\cos 0}{4}=\frac14+\frac14=\frac12$$
$$\int\limits_{0}^{\pi/4}(\cos^2 x-\sin^2 x)\,dx=\int\limits_{0}^{\pi/4}\cos 2x\,dx$$
$$=\frac12\sin 2x\Bigg|_{0}^{\pi/4}=\frac12\sin\frac{\pi}{2}-\frac12\sin 0=\frac12$$
$$\int\limits_{0}^{\pi}(\sin^4 x+\cos^4 x)\,dx$$
$$=\int\limits_{0}^{\pi}\left((\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x\right)\,dx$$
$$=\int\limits_{0}^{\pi}\left(1-\frac12\sin^2 2x\right)\,dx=\int\limits_{0}^{\pi}\left(1-\frac{1-\cos 4x}{4}\right)\,dx$$
$$=\int\limits_{0}^{\pi}\left(\frac34+\frac{\cos 4x}{4}\right)\,dx=\left(\frac{3x}{4}+\frac{\sin 4x}{16}\right)\Bigg|_{0}^{\pi}=\frac{3\pi}{4}$$
$$\int\limits_{0}^{3}x^2\sqrt{x+1}\,dx$$
Сделаем замену: $$t=x+1$$, тогда $$x=t-1$$, $$dx=dt$$, пределы: при $$x=0$$ имеем $$t=1$$, при $$x=3$$ имеем $$t=4$$.
$$\int\limits_{1}^{4}(t-1)^2\sqrt{t}\,dt=\int\limits_{1}^{4}(t^2-2t+1)t^{1/2}\,dt$$
$$=\int\limits_{1}^{4}\left(t^{5/2}-2t^{3/2}+t^{1/2}\right)\,dt$$
$$=\left(\frac{2}{7}t^{7/2}-\frac{4}{5}t^{5/2}+\frac{2}{3}t^{3/2}\right)\Bigg|_{1}^{4}$$
$$=\left(\frac{2}{7}\cdot 4^{7/2}-\frac{4}{5}\cdot 4^{5/2}+\frac{2}{3}\cdot 4^{3/2}\right)-\left(\frac{2}{7}-\frac{4}{5}+\frac{2}{3}\right)=\frac{586}{105}$$
$$\int\limits_{3}^{4}\frac{x^2-4x+5}{x-2}\,dx=\int\limits_{3}^{4}\left(x-2+\frac{1}{x-2}\right)\,dx$$
$$=\left(\frac{x^2}{2}-2x+\ln(x-2)\right)\Bigg|_{3}^{4}$$
$$=\left(8-8+\ln 2\right)-\left(\frac{9}{2}-6+\ln 1\right)=\frac32+\ln 2$$