Question

1)вычислить интегралы а) |(4/3х^3-3/2х^2+8)dx б) | sin^2xcosx dx

Answer 1

$\int\limits {( \frac{4}{3 x^{3} }- \frac{3}{2 x^{2} } +8) } \, dx = -\frac{2}{3 x^{2} }+ \frac{3}{2x}+8x+c$

$\int\limits { sin^{2} x}cosx \, dx= \int\limits ({ cos^{2}x- cos^{4}x }) \, dx = <img src=$- \frac{3x}{8} - \frac{3}{16} sin2x+c" alt=" \int\limits { cos^{2}x } \, dx =\frac{x}{2} + \frac{1}{4}sin2x- \frac{ cos^{2} xsinx}{4}- \frac{3}{4}" />- \frac{3x}{8} - \frac{3}{16} sin2x+c" />