Question

sin^4 2xdx интеграл​

Answer 1

$\displaystyle \int \sin^42xdx=\int\left(\dfrac{1-\cos4x}{2}\right)^2dx=\int\dfrac{1-2\cos 4x+\cos^24x}{4}dx=\\ \\ \\ =\dfrac{1}{4}\int dx-\dfrac{1}{2}\int\cos 4xdx+\dfrac{1}{8}\int \left(1+\cos 8x\right)dx=\\ \\ \\ =\dfrac{x}{4}-\dfrac{1}{8}\sin4x+\dfrac{x}{8}+\dfrac{1}{64}\sin 8x+C=\dfrac{3x}{8}-\dfrac{1}{8}\sin4x+\dfrac{1}{64}\sin8x+C$