Question

Вычислите интеграл
Варианты ответов
1) (π - 8) / 32
2) (3π - 4) / 32
3) (3π - 8) / 32

Answer 1

Ответ:   №3 .

$\int sin^4x\, dx=\int (sin^2x)^2\, dx=\int \Big(\dfrac{1-cos2x}{2}\Big)^2\, dx=\\\\=\dfrac{1}{4}\int (1-2cos2x+cos^22x)\, dx=\dfrac{1}{4}\int \Big(1-2\, cos2x+\dfrac{1+cos4x}{2}\Big)\, dx=\\\\=\dfrac{1}{4}\int \Big(\dfrac{3}{2}-2\, cos2x+\dfrac{1}{2}\, cos4x\Big)\, dx=\dfrac{1}{4}\cdot \Big(\dfrac{3}{2}\, x-sin2x+\dfrac{1}{8}\, sin4x\Big)+C=\\\\=\dfrac{3}{8}\, x-\dfrac{1}{4}\, sin2x+\dfrac{1}{32}\, sin4x+C$

$\int \limits _0^{\pi /4}sin^4x\, dx=\Big(\dfrac{3}{8}\, x-\dfrac{1}{4}\, sin2x+\dfrac{1}{32}\, sin4x\Big)\Big|_0^{\pi /4}=\\\\\\=\dfrac{3}{8}\cdot \dfrac{\pi}{4}-\dfrac{1}{4}\, sin\dfrac{\pi}{2}+\dfrac{1}{32}\, sin\pi =\dfrac{3\pi }{32}-\dfrac{1}{4}=\dfrac{3\pi -8}{32}$