Четврти разред средње школе

Неодређени интеграли 3

Интеграли. Особине интеграла. Примери табличних интеграла Једноставни примери.

Задаци

Текст задатака објашњених у видео лекцији:

Решити

Пр.13)    $\int {{2^x}} {e^x}dx$

Пр.14)   $\int {\frac{{3 \cdot {2^x} - 2 \cdot {3^x}}}{{2x}}} dx$

Пр.15)   $\int {\frac{{\cos 2x}}{{{{\sin }^2}x \cdot {{\cos }^2}x}}} dx$

Пр.16)   $\int {t{g^2}} xdx$

Пр.17)   $\int {\frac{{1 \cdot dx}}{{{{\sin }^2}x \cdot {{\cos }^2}x}}} $

Пр.18)   $\int {{{\cos }^2}} \frac{x}{2}dx$

 

Пр.13)    $ \int {{2^x}} {e^x}dx = \int {{{\left( {2e} \right)}^x}} dx = \frac{{{{\left( {2e} \right)}^x}}}{{\ln 2e}} + C = \frac{{{{\left( {2e} \right)}^x}}}{{\ln 2 + \ln e}} + C = \frac{{{{\left( {2e} \right)}^x}}}{{\ln 2}} + C$

 

Пр.14)   $\int {\frac{{3 \cdot {2^x} - 2 \cdot {3^x}}}{{{2^x}}}} dx = \int {\left( {\frac{{3 \cdot {2^x}}}{{{2^x}}} - \frac{{2 \cdot {3^x}}}{{{2^x}}}} \right)} dx = \int 3 dx - \int {\frac{{2 \cdot {3^x}}}{{{2^x}}}} dx = 3x - 2{\int {\left( {\frac{3}{2}} \right)} ^x}dx = $

$ = 3x - 2\frac{{{{\left( {\frac{3}{2}} \right)}^x}}}{{\ln \frac{3}{2}}} + C = 3x - 2\frac{{{{\left( {\frac{3}{2}} \right)}^x}}}{{\ln 3 - \ln 2}} + C$

 

Пр.15)   $\int {\frac{{\cos 2x}}{{{{\sin }^2}x \cdot {{\cos }^2}x}}} dx = \int {\frac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\sin }^2}x \cdot {{\cos }^2}x}}} dx = \int {\frac{{{{\cos }^2}x}}{{{{\sin }^2}x \cdot {{\cos }^2}x}}} dx - \int {\frac{{{{\sin }^2}x}}{{{{\sin }^2}x \cdot {{\cos }^2}x}}} dx = $

$ = \int {\frac{1}{{{{\sin }^2}x}}} dx - \int {\frac{1}{{{{\cos }^2}x}}} dx =  - ctgx - tgx + C$

 

Пр.16)   $\int {t{g^2}} xdx = \int {\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} dx = \int {\frac{{1 - {{\cos }^2}x}}{{{{\cos }^2}x}}} dx = \int {\frac{1}{{{{\cos }^2}x}}} dx - \int {\frac{{{{\cos }^2}x}}{{{{\cos }^2}x}}} dx = $

$ = tgx - \int {dx = } tgx - x + C$

 

Пр.17)   $\int {\frac{{1 \cdot dx}}{{{{\sin }^2}x \cdot {{\cos }^2}x}}}  = \int {\frac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x \cdot {{\cos }^2}x}}} dx = \int {\frac{{{{\sin }^2}x}}{{{{\sin }^2}x \cdot {{\cos }^2}x}}} dx + \int {\frac{{{{\cos }^2}x}}{{{{\sin }^2}x \cdot {{\cos }^2}x}}} dx =  $

$ = \int {\frac{1}{{{{\cos }^2}x}}} dx + \int {\frac{1}{{{{\sin }^2}x}}} dx = tgx - ctgx + C$

 

Пр.18)   $\int {{{\cos }^2}} \frac{x}{2}dx = \int {\frac{{1 + \cos x}}{2}} dx = \int {\frac{1}{2}} dx + \int {\frac{{\cos x}}{2}} dx = \frac{1}{2}x + \frac{1}{2}\operatorname{sinx}  + c$