MX 2 (2 Viewers)

umm what · Apr 21, 2012

can someone show me the working of how do we get tan (theta) when we integrate sec^2 Theta

thanks

nightweaver066 · Apr 21, 2012

$\frac{d}{d\theta}(tan\theta) = sec^2\theta$

$Don't believe me?$

$\frac{d}{d\theta}(tan\theta) = \frac{d}{d\theta}(\frac{sin\theta}{cos\theta})$

$= \frac{cos^2\theta + sin^2\theta}{cos^2\theta}$

$= \frac{1}{cos^2\theta}$

$= sec^2\theta$

$\therefore \int \frac{d}{d\theta}(tan\theta) d\theta = \int sec^2\theta d\theta$

$\int sec^2\theta d\theta = tan\theta + c$

Remember they're reverse processes.

barbernator · Apr 21, 2012

$tan(\alpha )=y\\ \alpha =tan-1(y)\\ \delta \alpha /\delta y=1/1+y^2\\ \delta y/\delta \alpha =1+y^2\\ =1+tan^2(\alpha )\\ =sec^2\alpha$

dayum

funnytomato · Apr 21, 2012

you get sec^2 when you differentiate tan
hence you get tan when you integrate sec^2

barbernator · Apr 21, 2012

funnytomato said:
you get sec^2 when you differentiate tan
hence you get tan when you integrate sec^2

u need to prove that homeboy

funnytomato · Apr 21, 2012

nightweaver066 said:
$\frac{d}{d\theta}(tan\theta) = sec^2\theta$

$Don't believe me?$

$\frac{d}{d\theta}(tan\theta) = \frac{d}{d\theta}(\frac{sin\theta}{cos\theta})$

$= \frac{cos^2\theta + sin^2\theta}{cos^2\theta}$

$= \frac{1}{cos^2\theta}$

$= sec^2\theta$

$\therefore \int \frac{d}{d\theta}(tan\theta) d\theta = \int sec^2\theta d\theta$

$\int sec^2\theta d\theta = tan\theta + c$

Remember they're reverse processes.

nice

funnytomato · Apr 21, 2012

barbernator said:
$tan(\alpha )=y\\ \alpha =tan-1(y)\\ \delta \alpha /\delta y=1/1+y^2\\ \delta y/\delta \alpha =1+y^2\\ =1+tan^2(\alpha )\\ =sec^2\alpha$

dayum

how did you get da/dy= 1/(1+y^2) ?

nightweaver066 · Apr 21, 2012

funnytomato said:
how did you get da/dy= 1/(1+y^2) ?

I believe he meant to type arctany, then differentiated that.

umm what · Apr 21, 2012

complicated much

barbernator said:
$tan(\alpha )=y\\ \alpha =tan-1(y)\\ \delta \alpha /\delta y=1/1+y^2\\ \delta y/\delta \alpha =1+y^2\\ =1+tan^2(\alpha )\\ =sec^2\alpha$

dayum

AAEldar · Apr 21, 2012

Here's the integral of $cosec^2x$ that I posted on here a while ago, it should be similar for $sec^2x$ . It's unmotivated in parts.

http://i.imgur.com/FblST.jpg

barbernator · Apr 21, 2012

raufmalik said:
complicated much

i don't find it complicated at all haha

funnytomato · Apr 21, 2012

nightweaver066 said:
I believe he meant to type arctany, then differentiated that.

yeah
but how do we know that the derivative of arctany= 1/(1+y^2)?

deswa1 · Apr 21, 2012

funnytomato said:
yeah
but how do we know that the derivative of arctany= 1/(1+y^2)?

This is a good point. I don't know why you'd try and establish that the integral of sec^2(x) is tanx. Its way easier for questions of this sort to establish that the derivative of tanx is sec^2(x) as above posters have done.

barbernator · Apr 21, 2012

funnytomato said:
yeah
but how do we know that the derivative of arctany= 1/(1+y^2)?

i guess i had kind of used a fact to prove a fact. should have done it the other way like nightweaver

funnytomato · Apr 21, 2012

barbernator said:
i guess i had kind of used a fact to prove a fact. should have done it the other way like nightweaver

repped

barbernator · Apr 21, 2012

funnytomato said:
repped

love it

get repped for fucking up ahah

AAEldar · Apr 21, 2012

$$Let $f(x)=\tan^{-1}x \\ $Then $f(\tan\theta)=\theta \\ \frac{d}{d\theta}f(\tan\theta)=\frac{d}{d\theta}\theta \\ f'(\tan\theta)\sec^2\theta=1 \\ f'(\tan\theta)=\frac{1}{\sec^2\theta} \\ $Since $\sec^2\theta=1+\tan^2\theta \\ f'(\tan\theta)=\frac{1}{1+\tan^2\theta} \\ $Since $x=\tan\theta \\ f'(x)=\frac{1}{1+x^2} \\ $Hence $\frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^2}$

There's some justification for you.

funnytomato · Apr 21, 2012

barbernator said:
love it get repped for fucking up ahah

rep >> everything else

btw, is there any particular reason that you're using that partial differentiation 'd' , instead of just dy/dx kind of d?

barbernator · Apr 21, 2012

the <a href="http://www.codecogs.com/eqnedit.php?latex=\delta" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\delta" title="\delta" /></a> is just delta which means "change in" in greek. I thought it was the "correct" letter to be used and people just used d for ease? or am i wrong?

this is the partial derivative symbol <a href="http://www.codecogs.com/eqnedit.php?latex=\partial" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\partial" title="\partial" /></a><a href="http://www.codecogs.com/eqnedit.php?latex=\partial" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\partial" title="\partial" /></a>

umm what · Apr 21, 2012

how is the integration of sex^2 (x/2) = 2 tan (x/2)

where does the two at front come from?? :S

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