Tìm: a) Nguyên hàm của 4(2-3x)^2 - 3 cos x

a) \(\int {\left[ {4{{\left( {2 - 3x} \right)}^2} - 3\cos x} \right]dx} \)\( = 4\int {{{\left( {2 - 3x} \right)}^2}dx} - 3\int {\cos xdx} \)

\( = 4\int {\left( {4 - 12x + 9{x^2}} \right)dx} - 3\int {\cos xdx} \)\( = 4\left( {4x - 6{x^2} + 3{x^3}} \right) - 3\sin x + C\)

= 16x – 24x² + 12x³ – 3sinx + C.

b) \(\int {\left( {3{x^3} - \frac{1}{{2{x^3}}}} \right)dx} \)\( = 3\int {{x^3}dx} - \frac{1}{2}\int {{x^{ - 3}}dx} \)\( = \frac{{3{x^4}}}{4} + \frac{1}{{4{x^2}}} + C\).

c) \(\int {\left( {\frac{2}{{{{\sin }^2}x}} - \frac{1}{{3{{\cos }^2}x}}} \right)dx} \)\( = 2\int {\frac{1}{{{{\sin }^2}x}}dx} - \frac{1}{3}\int {\frac{1}{{{{\cos }^2}x}}dx} \)\( = - 2\cot x - \frac{1}{3}\tan x + C\).

d) \(\int {\left( {{3^{2x - 2}} + 4\cos x} \right)dx} \)\( = \frac{1}{9}\int {{9^x}dx} + 4\int {\cos xdx} \)\( = \frac{1}{9}.\frac{{{9^x}}}{{\ln 9}} + 4\sin x + C\).

e) \(\int {\left( {4\sqrt[5]{{{x^4}}} + \frac{3}{{\sqrt {{x^3}} }}} \right)dx} \)\( = 4\int {{x^{\frac{4}{5}}}dx} + 3\int {{x^{\frac{{ - 3}}{2}}}dx} \)\( = \frac{{20}}{9}{x^{\frac{9}{5}}} - 6{x^{\frac{{ - 1}}{2}}} + C\).

g) \(\int {{{\left( {\sin \frac{x}{2} - \cos \frac{x}{2}} \right)}^2}dx} \)\( = \int {\left( {1 - 2\sin \frac{x}{2}\cos \frac{x}{2}} \right)dx} \)\( = \int {\left( {1 - \sin x} \right)dx} \)

\( = \int {dx} - \int {\sin xdx} \) = x + cosx + C.