Đáp án: $\frac{1}{16}sin{4x}$
Giải thích các bước giải:
$\int (sin^{4}x-cos^{4}x)dx=\int sin^{4}xdx - \int cos^{4}xdx=\int(sin^{2}x)^{2}dx -\int(cos^{2}x)^{2}dx
=\int(\frac{1}{2}(1-cos{2x}))^{2}dx - \int(\frac{1}{2}(cos{2x}+1))^{2}dx= =\frac{1}{4}\int(1-cos{2x})^{2}dx - \frac{1}{4}\int(cos{2x}
+1)^{2}dx= \frac{1}{4}\int(1-2cos{2x}+cos{2x}^{2})dx - \frac{1}{4}\int(1+2cos{2x}+cos{2x}^{2})dx= \frac{1}{4}\int(1-2cos{2x}+\frac{1}{2}+\frac{1}{2}cos{4x})dx- \frac{1}{4}\int(1+2cos{2x}+\frac{1}{2}+\frac{1}{2}cos{4x})dx=\frac{1}{4}\int(\frac{3}{2}-2cos{2x}+\frac{1}{2}cos{4x})dx - \frac{1}{4}\int(\frac{3}{2}+2cos{2x}+\frac{1}{2}cos{4x})dx=\frac{1}{4}(\frac{3}{2}x-sin{2x}+\frac{1}{8}sin{4x})-\frac{1}{4}(\frac{3}{2}x+sin{2x}+\frac{1}{8}sin{4x})=\frac{3}{8}x-\frac{1}{4}sin{2x}+\frac{1}{32}sin{4x}-\frac{3}{8}x+\frac{1}{4}sin{2x}+\frac{1}{32}sin{4x}=\frac{1}{16}sin{4x}$
