Question

найти интеграл sin^2x*cos^2x dx

Answer 1

$Sin^2xcdot Cos^2x=(Sinxcdot Cosx)(Sinxcdot Cosx)=frac{1}{4}(Sin2x)(Sin2x)= \ =frac{1}{4}Sin^22x \ Cos^2alpha -Sin^2alpha=Cos2alpha => 1-2Sin^2alpha=Cos2alpha => \ => Sin^2alpha=frac{1+Cos2alpha}{2} \ int {Sin^2xcdot Cos^2x} , dx =frac{1}{8}int {(1+Cos4x)dx=frac{1}{8}( int{}dx+int{Cos4x}dx)$
$int {}dx=x \ int {Cos4x} dx: \ 4x=u => 4xdx=du => frac{du}{4}=dx => \ => int{Cos4x}dx=frac{1}{4}int{Cosu}du=frac{Sinu}{4}=frac{Sin4x}{4} \ int {Cos4x} dx=frac{Sin4x}{4}$

$frac{1}{8}(int{}dx+int{Cos4x}dx)=frac{1}{8}(x+frac{Sin4x}{4})$