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anonymous
 4 years 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?
anonymous
 4 years 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?

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Valpey
 4 years ago
Best ResponseYou've already chosen the best response.1Asking cos*sec^2, not cos/sec^2

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, sorry: Cosx/Cos^2x = 1/Cosx = Secx

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0Well actually you'd keep it as 1/Cox, and use integration by parts

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait a second \[\int cosx\frac 1{cos^2x}dx?\]

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry for the confusion.. Let's start over: Cosx*Sec^2x = Cosx/Cos^2x = 1/Cosx, since Secx = 1/Cosx

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\frac 1{cosx}dx\]

Valpey
 4 years ago
Best ResponseYou've already chosen the best response.1\[\int{\cos{x}*\frac{1}{1\sin^2{x}}}\ dx\]

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0Well 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0why did you change it? wouldn't it make more sense to cancel one cosine out?

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

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0Actually this would be the easiest option, imo: Cos^2(u) = (1/2)+(Cos(2u))/2

bahrom7893
 4 years ago
Best ResponseYou've already chosen the best response.0ok let Valpey, finish this, otherwise you'll get confused.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do I multiply the integral by \[\ frac {1+u^2}{1+u^2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac {1+u^2}{1+u^2}\]

experimentX
 4 years ago
Best 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0makes sense, thank you!

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

Valpey
 4 years ago
Best ResponseYou've already chosen the best response.1*\[\frac{1}{2}\Big(\log(1+\sin x)\log(1\sin x)\Big)\]

Valpey
 4 years ago
Best ResponseYou've already chosen the best response.1Oh 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.

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1it seems that everyone here has forgotten the simple trick to this integral...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0see there had to be a simpler way

TuringTest
 4 years ago
Best 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)

TuringTest
 4 years ago
Best 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:\[u=\sec x+\tan x\implies du=\sec^2x+\sec x\tan xdx\]\[\int\frac1udu=\lnu+C=\ln\sec x+\tan x+C\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1that is the standard approach for this wellknown integral

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but yes it makes sense now I see it.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1I 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 :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I shall memorize this integral =P (honestly I will) And as always, Thanks Turing!
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