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\[\int cosxsec^2xdx\]
I forgot how to do this integral manually
this is what wolf had to say
http://www.wolframalpha.com/input/?i=integral+of+cosxsec^2x
how how do I do it manually?
 one year ago
 one year ago
\[\int cosxsec^2xdx\] I forgot how to do this integral manually this is what wolf had to say http://www.wolframalpha.com/input/?i=integral+of+cosxsec^2x how how do I do it manually?
 one year ago
 one year ago

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ValpeyBest ResponseYou've already chosen the best response.1
Asking cos*sec^2, not cos/sec^2
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.0
Oh, sorry: Cosx/Cos^2x = 1/Cosx = Secx
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.0
Well actually you'd keep it as 1/Cox, and use integration by parts
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
wait a second \[\int cosx\frac 1{cos^2x}dx?\]
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.0
Sorry for the confusion.. Let's start over: Cosx*Sec^2x = Cosx/Cos^2x = 1/Cosx, since Secx = 1/Cosx
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\int\frac 1{cosx}dx\]
 one year ago

ValpeyBest ResponseYou've already chosen the best response.1
\[\int{\cos{x}*\frac{1}{1\sin^2{x}}}\ dx\]
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.0
Well that I just had memorized, but actually i bet you don't simplify it down to 1/Cos, I was probably wrong again, you keep it as Cos/Cos^2 and then do what Valpey told you to
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
why did you change it? wouldn't it make more sense to cancel one cosine out?
 one year ago

ValpeyBest ResponseYou've already chosen the best response.1
Let u = sin(x) du = cos(x)dx \[\int{\frac{du}{1u^2}}\]
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.0
Actually this would be the easiest option, imo: Cos^2(u) = (1/2)+(Cos(2u))/2
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.0
ok let Valpey, finish this, otherwise you'll get confused.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
do I multiply the integral by \[\ frac {1+u^2}{1+u^2}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\frac {1+u^2}{1+u^2}\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
\[ \int \sec x dx = \log \left  \sec x + \tan x\right  = \log \left \tan \left( {x \over 2} + {\pi \over 4}\right)\right  \] wikipedia says it's quite a problem http://en.wikipedia.org/wiki/Integral_of_the_secant_function
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
makes sense, thank you!
 one year ago

ValpeyBest ResponseYou've already chosen the best response.1
Oh, duh! My bad.\[\int\frac{1}{1u^2}du=\int\frac{1}{(1u)(1+u)}du=\frac{1}{2}\int\frac{1+1}{(1u)(1+u)}du\]\[=\frac{1}{2}\int\Big(\frac{1+u}{(1u)(1+u)}+\frac{1u}{(1u)(1+u)}\Big)du\]\[=\frac{1}{2}\int\frac{1+u}{(1u)(1+u)}du+\frac{1}{2}\int\frac{1u}{(1u)(1+u)}du=\frac{1}{2}\int\frac{1}{1u}du+\frac{1}{2}\int\frac{1}{1+u}du\] \[=\frac{1}{2}(\log(1u))+\frac{1}{2}(\log(1+u))\] \[\frac{1}{2}\Big(\log(1+\sin x)\log(1+\sin x)\Big)\]
 one year ago

ValpeyBest ResponseYou've already chosen the best response.1
*\[\frac{1}{2}\Big(\log(1+\sin x)\log(1\sin x)\Big)\]
 one year ago

ValpeyBest ResponseYou've already chosen the best response.1
Oh yeah, I need absolute values. \[\frac{1}{2}\Big(\log1+\sin x−\log1−\sin x\Big)\]Which is also\[\frac{1}{2}\log\frac{1+\sin x}{1−\sin x}\], but now I'm just cheating off Wikipedia.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
@MathSofiya still want help with\[\int\cos x\sec^2xdx\]?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
it seems that everyone here has forgotten the simple trick to this integral...
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
see there had to be a simpler way
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[\cos x\sec^2x=\cos x\cdot\frac1{\cos^2 x}=\frac1{\cos x}=\sec x\]so\[\int\cos x\sec^2 xdx=\int\sec xdx\]now the great trick for this integral: multiply by\[{\sec x+\tan x\over\sec x+\tan x}\]and watch what happens (yes I know the idea is not obvious, but once you see it try to remember it)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[\int\sec x\left({\sec x+\tan x\over\sec x+\tan x}\right)dx=\int{\sec^2x+\sec x\tan x\over\sec x+\tan x}dx\]magically the top is now the derivative of the bottom:\[u=\sec x+\tan x\implies du=\sec^2x+\sec x\tan xdx\]\[\int\frac1udu=\lnu+C=\ln\sec x+\tan x+C\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
that is the standard approach for this wellknown integral
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
oh I see. So is it just a matter of doing many integrals and then coming to this conclusion? I wouldn't have thought of that, and yet again I haven't reviewed my integrals much.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
but yes it makes sense now I see it.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
I don't think I would have thought of this had I not known this integral either, I'm sure whoever discovered this trick spent quite some time on it. Fortunately, most integral tricks are far more obvious, this is just one of the more esoteric, yet directly effective tricks. There are few integrals that require such a nonobvious algebraic manipulation, most are easier too see. That is why I say, memorize this one, it is a classic and will serve you in the future :)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I shall memorize this integral =P (honestly I will) And as always, Thanks Turing!
 one year ago
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