1) Ta có: $∫(sin^42x+cos^42x)dx=∫[(sin^22x+cos^22x)^2-2sin^22x.cos^22x]dx$
$= ∫(1- \frac{2sin^24x}{4})dx=∫(1+ \frac{cos8x -1}{4})dx$
$ =∫(\frac{cos8x}{4} +\frac{3}{4})dx$
$=\frac{sin8x}{32} + \frac{3x}{4} + c$
2) Ta có: $∫(sin^43x-cos^43x)dx=∫(sin^23x-cos^23x)(sin^23x+cos^23x)dx$
$= ∫(sin^23x-cos^23x)dx=∫(-cos6x)dx=\frac{-sin6x}{6} +c$
3) Ta có: $∫(sin^6x+cos^6x)dx = ∫[(sin^2x+cos^2x)^3-3sin^2xcos^2x(sin^2x+cos^2x)]dx$
$ = ∫(1-3sin^2xcos^2x)dx=∫(1-\frac{3sin^22x}{4})dx$
$ =∫(\frac{3cos4x}{8}+\frac{5}{8})dx = \frac{3sin4x}{32} + \frac{5x}{8} +c$
4) Ta có: $∫(sin^63x-cos^63x)dx=∫[(sin^23x-cos^23x)^3 + 3sin^23xcos^23x(sin^23x-cos^23x)]dx$
$ = ∫(-cos^26x-\frac{3sin^26x.cos6x}{4})dx=∫-(cos^26x+\frac{3sin^26x}{4}).cos6xdx$
Đặt sin6x=t ⇒ dt=6cos6xdx
⇒∫$(\frac{t^2}{4}-1).\frac{dt}{6}= \frac{t^3}{72}- \frac{t}{6}+c$
