Giải thích các bước giải:
Ta có:
$sin^4x.cos^2x=sin^2x.(sinx.cosx)^2=\frac{1-cos2x}{2}.(\frac{1}{2}sin2x)^2$
$=\frac{1}{8}sin^22x-\frac{1}{8}sin^22x.cos2x$
$=\frac{1}{8}.\frac{1-cos4x}{2}-\frac{1}{8}.sin^22x.cos2x$
=>$I=\frac{1}{16}$$\int\limits {} \, dx-\frac{1}{16}$ $\int\limits {cos4x} \, dx-\frac{1}{8}$ $\int\limits {sin^2x.cos2x} \, dx$
$=\frac{1}{16}x-\frac{1}{64}sin4x-\frac{1}{48}sin^32x+C$
