Question

Найти первообразную для функций

Answer 1

$\displaystyle \int (sinx*cosx)dx=\bugg|sinx=t; cosxdx=dt\bigg|=\int tdt=\frac{t^2}{2}+C=\\\\=\frac{1}{2}sin^2x+C$

$\displaystyle \int (x^{3/2}+\frac{3}{2}x) dx=\frac{x^{5/2}}{5/2}+\frac{3}{2}\frac{x^2}{2}+C =\frac{2}{5}\sqrt{x^5}+\frac{3}{4}x^2+C$