Đáp án:
Giải thích các bước giải:
$\begin{array}{l} \int {{{\sin }^6}x + co{s^6}x} dx\\ = \int {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x - {{\sin }^2}x.{{\cos }^2}x + {{\cos }^4}x} \right)dx} \\ = \int {1.\left( {{{\sin }^4}x + 2{{\sin }^2}{{\cos }^2}x + {{\cos }^4}x - 3{{\sin }^2}x.{{\cos }^2}x} \right)dx} \\ = \int {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - \frac{3}{4}.4{{\sin }^2}x.{{\cos }^2}x} dx\\ = \int {1 - \frac{3}{4}{{\sin }^2}2xdx} \\ = x - \int {\frac{3}{8}.2{{\sin }^2}2xdx} \\ = x - \frac{3}{8}\int {\left( {1 - \cos 4x} \right)dx} \\ = x - \frac{3}{8}\left( {x - \frac{1}{4}\sin 4x} \right) + C\\ = \frac{5}{8}x + \frac{1}{4}\sin 4x + C \end{array}$
