Question

Помогите дорешать интеграл. Пример: $\int\limits {sin^2x*cos^4x} \, dx = \int\limits {(sin^2x*cos^2x)cos^2x} \, dx = .....$. Ответ должен быть: $\frac{1}{48} sin^32x+\frac{1}{16}x-\frac{1}{64} sin4x+ C$

Answer 1

Ответ:

Пошаговое объяснение:

$\int {\sin^2x\cos^4x} \, dx =\int {(\sin^2x\cos^2x)\cos^2x} \, dx =\int {(\sin x\cos x)^2(\frac{1}{2}+\frac{\cos2x}{2} ) } \, dx =\\\int {\frac{1}{4} \sin^22x(\frac{1}{2}+\frac{\cos2x}{2} ) } \, dx = \frac{1}{4} \int {(\frac{1}{2}-\frac{\cos4x}{2} ) (\frac{1}{2}+\frac{\cos2x}{2} )} \, dx =\\\frac{1}{4} \int {(\frac{1}{4} +\frac{1}{4}\cos2x -\frac{1}{4}\cos4x-\frac{1}{4}\cos2x\cos4x) } \, dx =\\\frac{1}{16} \int {(1+\cos2x-\cos4x-\frac{1}{2}\cos2x -\frac{1}{2} \cos6x)} \, dx =$

$\frac{1}{16} \int {(1-\cos4x+\frac{1}{2}\cos2x -\frac{1}{2} \cos6x)} \, dx =\\\frac{1}{16}\int {} \, dx -\frac{1}{64}\int {\cos4x*} \, 4dx +\frac{1}{64}\int {\cos2x*} \, 2dx -\frac{1}{192}\int {\cos6x*} \, 6dx =\\\frac{1}{16}\int {} \, dx -\frac{1}{64}\int {\cos4x} \, d(4x) +\frac{1}{64}\int {\cos2x} \, d(2x) -\frac{1}{192}\int {\cos6x} \, d(6x)=\\\frac{1}{16}x+\frac{1}{64}\sin2x-\frac{1}{64}\sin4x-\frac{1}{192}\sin6x +C=$

$\frac{1}{48}(\frac{3\sin2x}{4} -\frac{\sin6x}{4} )+ \frac{1}{16}x-\frac{1}{64}\sin4x +C=\\\\\frac{1}{48}\sin^32x+\frac{1}{16}x-\frac{1}{64}\sin4x+C$