РЕШИТЕ , ОЧ НАДО $\int\limits^\pi _0{sinx^{4} } \, dx \\$

(там над интегралом пи/2, не получается знак деления поставить)
и там sin^4x

(там cos^4x)
dx/ $\sqrt{4-3x}$
dx/ $\sqrt{3x-2}$

$\int\limits^\pi_0 {cosx^{4} } \, dx$
$\int\limits^1_0$
$\int\limits^6_2$

Ответ:

DevilWoods

14.07.2021 09:31

$1)\ \ \displaystyle \int\limits_0^{\pi /2}\, sin^4x\, dx=\int\limits_0^{\pi /2}\, (sin^2x)^2\, dx=\int\limits_0^{\pi /2}\, \Big(\frac{1-cos2x}{2}\Big)^2\, dx=\\\\\\=\int\limits_0^{\pi /2}\, \frac{1-2\, cos2x+cos^22x}{4}\, dx=\frac{1}{4}\int\limits_0^{\pi /2}\, \Big(1-2\, cos2x+\frac{1+cos4x}{2}\Big)\, dx=\\\\\\=\frac{1}{4}\int\limits_0^{\pi /2}\, \Big(1-2\, cos2x+\frac{1}{2}+\frac{1}{2}cos4x\Big)\, dx=\frac{1}{4}\int\limits_0^{\pi /2}\, \Big(\frac{3}{2}-2\, cos2x+\frac{1}{2}cos4x\Big)\, dx=$

$\displaystyle =\frac{1}{4}\cdot \Big(\frac{3}{2}\, x-sin2x+\frac{1}{8}\, sin4x \Big)\Big|_0^{\pi /2}=\frac{1}{4}\cdot \Big(\frac{3\pi}{4}-0+0\Big)=\frac{3\pi }{16}$

$2)\ \ \displaystyle \int\limits_0^{\pi /2}\, cos^4x\, dx=\int\limits_0^{\pi /2}\, (cos^2x)^2\, dx=\int\limits_0^{\pi /2}\, \Big(\frac{1+cos2x}{2}\Big)^2\, dx=\\\\\\=\frac{1}{4}\int\limits_0^{\pi /2}\, \Big(1+2\, cos2x+cos^22x\Big)\, dx=\frac{1}{4}\int\limits_0^{\pi /2}\, \Big(1+2\, cos2x+\frac{1+cos4x}{2}\Big)\, dx=\\\\\\=\frac{1}{4}\int\limits_0^{\pi /2}\, \Big(1+2\, cos2x+\frac{1}{2}+\frac{1}{2}cos4x\Big)\, dx=\frac{1}{4}\int\limits_0^{\pi /2}\, \Big(\frac{3}{2}+2\, cos2x+\frac{1}{2}cos4x\Big)\, dx=$

$\displaystyle =\frac{1}{4}\cdot \Big(\frac{3}{2}\, x+sin2x+\frac{1}{8}\, sin4x \Big)\Big|_0^{\pi /2}=\frac{1}{4}\cdot \Big(\frac{3\pi}{4}+0+0\Big)=\frac{3\pi }{16}$

$\star \ \ \displaystyle \int\limits_0^{\pi }\, cos^4x\, dx=\frac{1}{4}\cdot \Big(\frac{3}{2}\, x+sin2x+\frac{1}{8}\, sin4x \Big)\Big|_0^{\pi }=\frac{1}{4}\cdot \Big(\frac{3\pi}{2}+0+0\Big)=\frac{3\pi }{8}\ \ \star$

$3)\ \ \displaystyle \int\limits^1_0\, \sqrt{4-3x}\, dx=-\frac{1}{3}\cdot \frac{(4-3x)^{3/2}}{3/2}\Big|_0^1=-\frac{2\sqrt{(4-3x)^3}}{9}\Big|_0^1=\\\\=-\frac{2}{9}\cdot \Big(\sqrt{1^3}-\sqrt{4^3}\Big)=-\frac{2}{9}\cdot (1-2^3)=-\frac{2}{9}\cdot (1-8)=\frac{14}{9}$

$4)\ \ \displaystyle \int\limits_2^6\, \frac{dx}{\sqrt{3x-2}}=\frac{1}{3}\cdot 2\sqrt{3x-2}\Big|_2^6=\frac{2}{3}\cdot (\sqrt{16}-\sqrt{4})=\frac{2}{3}\cdot (4-2)=\frac{4}{3}$