Question

Tìm:

a) \(\int\left[4\left(2-3x\right)^2-3\cos x\right]dx;\)

b) \(\int\left(3x^3-\dfrac{1}{2x^3}\right)dx;\)

c) \(\int\left(\dfrac{2}{\sin^2x}-\dfrac{1}{3\cos^2x}\right)dx;\)

d) \(\int\left(3^{2x-2}+4\cos x\right)dx;\)

e) \(\int\left(4\sqrt[5]{x^4}+\dfrac{3}{\sqrt{x^3}}\right)dx;\)

g) \(\int\left(\sin\dfrac{x}{2}-\cos\dfrac{x}{2}\right)^2dx\).

Answer 1

a) \(\int {\left[ {4{{\left( {2 - 3x} \right)}^2} - 3\cos x} \right]dx}  = 4\int {{{\left( {2 - 3x} \right)}^2}dx}  - 3\int {\cos xdx}  = 4\int {\left( {9{x^2} - 12x + 4} \right)dx}  - 3\int {\cos xdx} \)

\( = 4\left( {3{x^3} - 6{x^2} + 4x} \right) - 3\sin x + C = 12{x^3} - 24{x^2} + 16x - 3\sin x + C\)

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

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

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

\( = \frac{1}{9}.\frac{{{9^x}}}{{\ln 9}} + 4\sin x + C = \frac{{{9^{x - 1}}}}{{\ln 9}} + 4\sin x + C\)

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

\( = 20\sqrt[5]{x} - \frac{6}{{{x^{\frac{1}{2}}}}} + C = 20\sqrt[5]{x} - \frac{6}{{\sqrt x }} + C\)

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

\( = \int {\left( {1 - \sin x} \right)dx}  = x - \left( { - \cos x} \right) + C = x + \cos x + C\)