Question

Integral

Answer 1

\[\frac{ 1 }{ 2 }\int\limits \tan2xdx\]

Answer 2

is it \[-\frac{ 1 }{ 4 }\ln \left| \cos2x \right|+C\]

Answer 3

\(\Huge \checkmark\)

Answer 4

seems good to me too... even though the book says sec2x instead of cos2x

Answer 5

do they give that answer but without the negative in front?

Answer 6

nope, it is still negative :/

Answer 7

Hm that is a bit odd. Because
since \[ -\ln x=(-1)\ln x=\ln x^{-1}=\ln \left( \dfrac{1}{x}\right)\]
So I figured.. \[ -\ln(\cos x)=\ln[ (\cos x)^{-1}]=\ln\left(\frac{1}{\cos x} \right)=\ln(\sec x)\]

Answer 8

ohh I see... but thanks for your help :) helped me to clear a bit more

Answer 9

:) but yeah the negative sign still in front would be strange to me!

Answer 10

well... it is the second part from the formula int(udv) = uv -int(vdu) so it really starts out as a positive

Answer 11

oh um.. so you have \[ uv-\int v\, du\]
but your integral \(\int v \, du\) was \(\int\frac{1}{2} \tan (2x)\,dx\) ?

Answer 12

I have u=x du =1, dv = sec^2(2x) v = (tan2x/2)-1

Answer 13

\[x \frac{ \tan2x }{ 2 }-\int\limits \frac{ \tan2x }{ 2 }*1\]

Answer 14

should look like this right?

Answer 15

yes

Answer 16

oh ok yeah there will be a negative .. because there is a negative in front of the integral.

Answer 17

yep coming from the formula... but the book says it is a negative answer and gives a sec2x instead of cos2x :/

Answer 18

\[ \int\frac{1}{2}\tan(2x)\, dx=-\frac{ 1 }{ 4 }\ln \left| \cos2x \right|+C=\ln|\sec{2x}|+C\\
\\ ~ \\-\int\frac{1}{2}\tan(2x)\, dx=-\ln|\sec{2x}|+C\]

Answer 19

what happens to the 1/4 in the first part when you make cos a sec?

Answer 20

oh oops sorry I definitely forgot to re-write the 1/4 when I made it sec, but just keep it in front of the \(\ln\)

Answer 21

ohh cool Thanks for your help we finally got it :)

Answer 22

awesome!