Question

integral of 2-3sin2x / cos2x dx?

Answer 1

\[\int\limits{\frac{2-3\sin(2x)}{\cos(2x)}}dx\]??

Answer 2

yup. :)

Answer 3

have you tried separating the terms?

Answer 4

\[\int\limits\frac{2}{\cos2x} -\frac{3\sin2x}{\cos2x} dx => \int\limits2*\frac{1}{\cos2x} - 3\tan2x dx => \int\limits2\sec2x -3\tan2x\]

Answer 5

\[\int\limits\frac{2}{\cos2x} -\frac{3\sin2x}{\cos2x} dx => \int\limits2*\frac{1}{\cos2x} - 3\tan2x dx =>\]
\[ \int\limits2\sec2x -3\tan2x\]

Answer 6

\[\int\limits2\sec2x dx - \int\limits3\tan2x dx\]

Answer 7

integrate term by term

Answer 8

the answer is
\[\ln |(1+\sin2x)|+1/2\ln|\cos2x|+C\]
can you show the solution?

Answer 9

Are you able to integrate the first one?

Answer 10

nope @Mimi_x3

Answer 11

can you integrate that term @Mimi_x3. will you please show me the integration. :)

Answer 12

\[\int\limits{\frac{2}{\cos(2x)}}dx=\int\limits{2\sec(2x)}dx\]let u=2x
du=2dx
\[\int\limits{\sec(u)du}\]\[=\int\limits{secu \times \frac{secu+tanu}{secu+tanu}du}=\int\limits{\frac{secu \tan u + \sec^2u}{secu+tanu}}du\]let V=secu+tanu
dv=secutanu+sec^2u
so the integral becomes
\[\int\limits{\frac{dv}{v}}=\ln(v)=\ln(secu+tanu)\]
\[=\ln(\sec(2x)+\tan(2x))\]

for the second integral
\[\int\limits{\frac{\sin(2x)}{\cos(2x)}dx}\]let u=cos(2x)
du=-2sin(2x)dx\[\int\limits{\frac{-1}{2}\frac{du}{u}}=-\frac{1}{2}\ln(u)=-\frac{1}{2}\ln(\cos(2x)\]

Answer 13

im too slow :(

Answer 14

its ok @Mimi_x3 thanks for your effort. :)

Answer 15

also thanks for @lalaly . :)

Answer 16

entonces cual es el resultado?¿

Answer 17

solo veo una parte del resultado

Answer 18

int (2-3sin2x)/cos 2x