$\sin^6 x +\cos^6 x= (\sin^2 x +\cos^2 x)(\cos^4 x -\sin^2 x \cos^2 x+\cos^4 x) = (\cos^2 x +\sin^2 x)^2 -3\sin^2 x \cos^2 x$$=1-\dfrac{3}{4}\sin^2 2x =1-\dfrac{3}{8}(1-\cos 4x) = \dfrac{5}{8}+\dfrac{3}{8}\cos 4x$Vậy $I = \int (\dfrac{5}{8}+\dfrac{3}{8}\cos 4x) dx = \dfrac{5}{8}\int dx + \dfrac{3}{8} \int \cos 4x dx = \dfrac{5}{8}x +\dfrac{3}{32} \int \cos 4x d(4x)$$= \dfrac{5}{8}x +\dfrac{3}{32} \sin 4x + C$ tự thế cận
$\sin^6 x +\cos^6 x= (\sin^2 x +\cos^2 x)(\cos^4 x -\sin^2 x \cos^2 x+\cos^4 x) = (\cos^2 x +\sin^2 x)^2 -3\sin^2 x \cos^2 x$$=1-\dfrac{3}{2}\sin^2 2x =1-\dfrac{3}{4}(1-\cos 4x) = \dfrac{1}{4}+\dfrac{3}{4}\cos 4x$Vậy $I = \int (\dfrac{1}{4}+\dfrac{3}{4}\cos 4x) dx = \dfrac{1}{4}\int dx + \dfrac{3}{4} \int \cos 4x dx = \dfrac{1}{4}x +\dfrac{3}{16} \int \cos 4x d(4x)$$= \dfrac{1}{4}x +\dfrac{3}{16} \sin 4x + C$ tự thế cận
$\sin^6 x +\cos^6 x= (\sin^2 x +\cos^2 x)(\cos^4 x -\sin^2 x \cos^2 x+\cos^4 x) = (\cos^2 x +\sin^2 x)^2 -3\sin^2 x \cos^2 x$$=1-\dfrac{3}{
4}\sin^2 2x =1-\dfrac{3}{
8}(1-\cos 4x) = \dfrac{
5}{
8}+\dfrac{3}{
8}\cos 4x$Vậy $I = \int (\dfrac{
5}{
8}+\dfrac{3}{
8}\cos 4x) dx = \dfrac{
5}{
8}\int dx + \dfrac{3}{
8} \int \cos 4x dx = \dfrac{
5}{
8}x +\dfrac{3}{
32} \int \cos 4x d(4x)$$= \dfrac{
5}{
8}x +\dfrac{3}{
32} \sin 4x + C$ tự thế cận