Задача 70169 ...

641174128cfb9210f616c190

16.03.2023 23:56:21

Вычислить интеграл ∫ sin^4xcos^2x dx

5f3ea7e3faf909182968ddd9

17.03.2023 00:17:46

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[m] ∫ (\frac{1-cos2x}{2})^2\cdot \frac{1+cos2x}{2}dx=[/m]


[m]= ∫ \frac{1-2cos2x+cos^22x}{4}\cdot \frac{1+cos2x}{2}dx=[/m]

[m] =\frac{1}{8}∫ (1-2cos2x+cos^22x)\cdot (1+cos2x)dx=[/m]

[m] =\frac{1}{8}∫ (1-2cos2x+cos^22x +cos2x-2cos^22x+cos^32x)dx=[/m]

[m] =\frac{1}{8}∫ (1-cos2x -cos^22x+cos^32x)dx=[/m]

[m] =\frac{1}{8}∫ (1-cos2x -\frac{1+cos4x}{2}+cos^22x\cdot cos2x)dx=[/m]

[m] =\frac{1}{8}∫ (\frac{1}{2}-cos2x -\frac{1}{2}cos4x +(1-sin^22x)\cdot cos2x)dx=[/m]

[m] =\frac{1}{8}∫ (\frac{1}{2}-\frac{1}{2}cos4x -sin^22x\cdot cos2x)dx=[/m]

[m]=\frac{1}{8}(\frac{1}{2}x-\frac{1}{8}sin4x-\frac{1}{2}\frac{sin^32x}{3})+C[/m]