Question

Evaluate: $\int \frac{{\sin }^{6}x+{\cos }^{6}x}{{\sin }^{2}x+{\cos }^{2}x}dx$

Answer 1

$I=\int \frac{{\sin }^{6}x+{\cos }^{6}x}{{\sin }^{2}x+{\cos }^{2}x}dx$

$=\int \frac{({\sin }^{2}x+{\cos }^{2}x)({\sin }^{4}x+{\cos }^{4}x-{\sin }^{2}x{\cos }^{2}x)}{({\sin }^{2}x+{\cos }^{2}x)}$

$=\int ({\sin }^{2}x+{\cos }^{2}x)-3{\sin }^{2}x{\cos }^{2}xdx$

$=\int 1-\frac{3}{4}{\sin }^{2}2xdx$

$=\int 1-\frac{3}{8}(1-\cos 4x)dx$

$=\int (\frac{5}{8}+\frac{3}{8}\cos 4x)dx=\frac{5x}{8}+\frac{3}{32}\sin 4x+c$