Question

интегрирование некоторых ....​

Answer 1

Ответ:

Объяснение:

1.

$a)\ \int\limits {(3tgx+\frac{5}{cos^2x} -3)} \, dx=3*\int\limits {tgx} \, dx +5*\int\limits {\frac{1}{cos^2x} } \, dx -3*\int\limits \, dx\\ \int\limits {tgx} \, dx =\int\limits {\frac{sinx}{cosx} } \, dx=\left | {{u=cosx} \atop {du=-sinxdx}} \right. |=-\int\limits \frac{du}{u} =-ln|u|+C=-ln|cosx|+C.\\\int\limits {\frac{1}{cos^2x} } \, dx=tgx+C.\\ \int\limits\, dx =x+C.\ \ \ \ \Rightarrow\\3*(-ln|cosx|)+5*tgx-3x+C=-3ln|cosx|+5tgx-3x+C.$

$b)\ \int\limits {(1+cos x)^2} \, dx =\int\limits {(1+2*cosx*1+cos^2x)} \, dx=\\=\int\limits\, dx +2*\int\limits {cosx} \, dx +\int\limits {cos^2x} \, dx =\\=x+2sinx+\int\limits {\frac{1+cos2x}{2} } \, dx +x+2sinx+\int\limits {(\frac{1}{2}+\frac{cos2x}{2} )} \, dx=\\=x+2sinx+\frac{1}{2} *\int\limits \, dx +\int\limits\frac{cos2x}{2} dx=x+2sinx+\frac{x}{2}+ \frac{sin2x}{4}+C=\\=\frac{3x}{2} +2sinx+\frac{sin2x}{4}+C.$

$c)\ \int\limits {\frac{2}{sin^2xcos^2x} } \, dx =2*\int\limits \frac{1}{sin^2xcos^2x}dx=2*\int\limits {\frac{sin^2x+cos^2x}{sin^2xcos^2x} } \, dx=2*\int\limits {(\frac{1}{cos^2x} +\frac{1}{sin^2x}) } \, dx=\\ =2*\int\limits {\frac{1}{cos^2x} } \, dx +2*\int\limits {\frac{1}{sin^2x} } \, dx =2tgx -2ctgx.$

2.

$a)\ \int\limits {cos(9x-\frac{\pi }{18} }) \, dx =\left | {{u=9x+\frac{\pi }{18}\ \ \ \ du=9dx } \atop {dx=\frac{du}{9} }} \right. |=\frac{1}{9} *\int\limits {cosu} \, du =\frac{sinu}{9}+C=\\ =\frac{sin(9x-\frac{\pi }{18} )}{9}+C .$

$b)\ \int\limits8^{2x-7} \, dx =\left | {{u=2x-7\ \ \ \ \ du=2dx}}\atop {dx=\frac{du}{2} }} \right. |=\frac{1}{2}* \int\limits {8^u} \, du =\frac{8^u}{2*ln8}+C=\frac{8^{2x-7}}{2ln8} +C.$

3.

$\int\limits {(\sqrt{1-e^x}*e^x) } \, dx =\left | {{u=e^x} \atop {du=e^xdx}} \right. |=\int\limits {(\sqrt{1-u } )\, du= \left \{ {{t=1-u\ \ \ \ u=1-t} \atop {du=-dt}} \right.| =-\int\limits {\sqrt{t} } \, dt =$

$=-\frac{2}{3} *t^{\frac{3}{2}}+C=-\frac{2}{3}*t*\sqrt{t}+C=-\frac{2}{3} *(1-u)*\sqrt{1-u} +C=$

$=-\frac{2}{3}*\sqrt{1-u} +\frac{2}{3} *\sqrt{1-u}*u+C=\frac{2\sqrt{1-e^x}*e^x }{3} -\frac{2\sqrt{1-e^x} }{3}+C.$