Integral (1 Viewer)

fortune_cookie · May 28, 2012

= tan(x) / [ 2cos^2(x) -1]

= sin(x)/ {cos(x)[2cos^2(x)-1]}

u= cos(x)

du=-sin(x) dx

I = -int 1/ [u(2u^2 -1) ] du

Partial fractions etc...

= int [ (1/u) - (2u)/(2u^2-1) ] du

= ln|u| - (1/2) ln|2u^2-1| +C

Back sub

ln|cos(x)| - (1/2)ln|cos(2x)| +C

barbernator · May 28, 2012

-( <a href="http://www.codecogs.com/eqnedit.php?latex=-ln(cos(x)) @plus; \frac{ln(cos^2(x)-\frac{1}{2})}{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?-ln(cos(x)) + \frac{ln(cos^2(x)-\frac{1}{2})}{2}" title="-ln(cos(x)) + \frac{ln(cos^2(x)-\frac{1}{2})}{2}" /></a> )+C

yay or nay?

SpiralFlex · May 28, 2012

barbernator said:
<a href="http://www.codecogs.com/eqnedit.php?latex=-ln(cos(x)) @plus; \frac{ln(cos^2(x)-\frac{1}{2})}{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?-ln(cos(x)) + \frac{ln(cos^2(x)-\frac{1}{2})}{2}" title="-ln(cos(x)) + \frac{ln(cos^2(x)-\frac{1}{2})}{2}" /></a>

yay or nay?

nay +C

fortune_cookie · May 28, 2012

barbernator said:
<a href="http://www.codecogs.com/eqnedit.php?latex=-ln(cos(x)) @plus; \frac{ln(cos^2(x)-\frac{1}{2})}{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?-ln(cos(x)) + \frac{ln(cos^2(x)-\frac{1}{2})}{2}" title="-ln(cos(x)) + \frac{ln(cos^2(x)-\frac{1}{2})}{2}" /></a>

yay or nay?

nay, it's close though.

I got wolframa to differentiate what you got and it is only out a negative sign.

i.e. the derivative of your answer is -tan(x)sec(2x).

What method did you use?

barbernator · May 28, 2012

fortune_cookie said:
nay, it's close though.

I got wolframa to differentiate what you got and it is only out a negative sign.

i.e. the derivative of your answer is -tan(x)sec(2x).

What method did you use?

same as urs until i got to partial fractions, where i split into linear factors so i could use the cover up method to evaluate. Must have changed a sign somewhere.

oh yeh, when i subbed in du=-sinxdx forgot to change sign

all good in da hood

Trebla · May 29, 2012

Another possible approach:

$Let $t=\tan x\Rightarrow dt = \sec^2x\,dx\Rightarrow dx = \dfrac{dt}{1+\tan^2x}\Rightarrow dx=\dfrac{dt}{1+t^2}\\\\$Note that $\sec 2x = \dfrac{1}{\cos 2x} = \dfrac{1+t^2}{1-t^2}\\\\\displaystyle\int\tan x\sec 2x\,dx\\\\= \displaystyle\int \dfrac{t(1+t^2)}{1-t^2}\times \dfrac{1}{1+t^2}\,dt\\\\=\displaystyle\int \dfrac{t}{1-t^2}\,dt \\\\=-\dfrac{1}{2}\ln |1-t^2| + c\\\\= -\dfrac{1}{2}\ln |1-\tan^2x| + c\\\\$Note that further simplification of this answer shows its equivalence to the answers shown in the above posts$

fortune_cookie · May 29, 2012

Trebla said:
Another possible approach:

$Let $t=\tan x\Rightarrow dt = \sec^2x\,dx\Rightarrow dx = \dfrac{dt}{1+\tan^2x}\Rightarrow dx=\dfrac{dt}{1+t^2}\\\\$Note that $\sec 2x = \dfrac{1}{\cos 2x} = \dfrac{1+t^2}{1-t^2}\\\\\displaystyle\int\tan x\sec 2x\,dx\\\\= \displaystyle\int \dfrac{t(1+t^2)}{1-t^2}\times \dfrac{1}{1+t^2}\,dt\\\\=\displaystyle\int \dfrac{t}{1-t^2}\,dt \\\\=-\dfrac{1}{2}\ln |1-t^2| + c\\\\= -\dfrac{1}{2}\ln |1-\tan^2x| + c\\\\$Note that further simplification of this answer shows its equivalence to the answers shown in the above posts$

Yeah, it looks like wolframa does something similiar to a t=tan(x/2) substitution except it is about 10 times longer and has a ton of unnessecary algebra and messy radicals everywhere lol: http://www.wolframalpha.com/input/?i=tanxsec(2x)

Stupid wolframa.

asianese · May 29, 2012

LOL THE WOLFRAM SOLUTION!! 'Simplify' => humongous denominator with polynomials degree 6...yep

juantheron · May 30, 2012

Thanks friends got it

although i have solved using Very Lengthy Method.

but Here Moderator solution is very short.

Thanks

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Integral (1 Viewer)

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