Question

Помогите буду благодарен.

Answer 1

$\int sin^4x\, dx=\int (sin^2x)^2\, dx=\int (\frac{1-cos2x}{2})^2\, dx=\frac{1}{4}\int (1-2cos2x+cos^22x)\, dx=\\\\=\frac{1}{4}\int (1-2cos2x+\frac{1+cos4x}{2})\, dx=\frac{1}{4}\cdot (x-\frac{2}{2}\, sin2x+\frac{1}{2}x+\frac{1}{2\cdot 4}sin4x)+C=\\\\=\frac{1}{4}\cdot (\frac{3}{2}x-sin2x+\frac{1}{8}sin4x)+C\\\\\\\int\limits^{\pi /2}_0\, sin^4x\, dx=\frac{1}{4}\cdot (\frac{3}{2}x-sin2x+\frac{1}{8}sin4x)\Big |_0^{\pi /2}=\frac{1}{4}\cdot \frac{3\pi }{4}=\frac{3\pi }{16}$

Answer 2

$\int\limits^\frac{\pi }{2} _0 {Sin^{4}} \, dx=\frac{1}{4}\int\limits^\frac{\pi }{2} _0 {(1-Cos2x)^{2}} \, dx=\frac{1}{4}\int\limits^\frac{\pi }{2} _0 {(1-2Cos2x+Cos^{2}2x) } \, dx=$$\frac{1}{4}\int\limits^\frac{\pi }{2} _0 {(1-2Cos2x+\frac{1}{2}+\frac{1}{2}Cos4x)} \, dx=\frac{1}{4} \int\limits^\frac{\pi }{2} _0 {(\frac{3}{2}-2Cos2x+\frac{1}{2}Cos4x)} \, dx=$$\frac{1}{4}\int\limits^\frac{\pi }{2} _0 {(\frac{3}{2}x-2*\frac{1}{2}Sin2x+\frac{Sin4x}{4*2})} \, dx =\frac{1}{4}{(\frac{3}{2}x-Sin2x+\frac{Sin4x}{8})}|^{\frac{\pi }{2}} _{0}=\frac{1}{4}(\frac{3}{2}*\frac{\pi }{2}-Sin\pi+\frac{Sin2\pi }{8}-\frac{3}{2}*0+Sin0-\frac{Sin0}{8})=\frac{1}{4}*\frac{3\pi }{4}=\frac{3\pi }{16}$