Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

\[\int cosxsec^2xdx\] 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?

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

Asking cos*sec^2, not cos/sec^2
Oh, sorry: Cosx/Cos^2x = 1/Cosx = Secx
Well actually you'd keep it as 1/Cox, and use integration by parts

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

wait a second \[\int cosx\frac 1{cos^2x}dx?\]
Yes.
Sorry for the confusion.. Let's start over: Cosx*Sec^2x = Cosx/Cos^2x = 1/Cosx, since Secx = 1/Cosx
\[\int\frac 1{cosx}dx\]
\[\int{\cos{x}*\frac{1}{1-\sin^2{x}}}\ dx\]
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
why did you change it? wouldn't it make more sense to cancel one cosine out?
No
Let u = sin(x) du = cos(x)dx \[\int{\frac{du}{1-u^2}}\]
Actually this would be the easiest option, imo: Cos^2(u) = (1/2)+(Cos(2u))/2
ok let Valpey, finish this, otherwise you'll get confused.
sure
do I multiply the integral by \[\ frac {1+u^2}{1+u^2}\]
\[\frac {1+u^2}{1+u^2}\]
\[ \int \sec x dx = \log \left | \sec x + \tan x\right | = \log \left| \tan \left( {x \over 2} + {\pi \over 4}\right)\right | \] wikipedia says it's quite a problem http://en.wikipedia.org/wiki/Integral_of_the_secant_function
makes sense, thank you!
Oh, duh! My bad.\[\int\frac{1}{1-u^2}du=\int\frac{1}{(1-u)(1+u)}du=\frac{1}{2}\int\frac{1+1}{(1-u)(1+u)}du\]\[=\frac{1}{2}\int\Big(\frac{1+u}{(1-u)(1+u)}+\frac{1-u}{(1-u)(1+u)}\Big)du\]\[=\frac{1}{2}\int\frac{1+u}{(1-u)(1+u)}du+\frac{1}{2}\int\frac{1-u}{(1-u)(1+u)}du=\frac{1}{2}\int\frac{1}{1-u}du+\frac{1}{2}\int\frac{1}{1+u}du\] \[=\frac{1}{2}(-\log(1-u))+\frac{1}{2}(\log(1+u))\] \[\frac{1}{2}\Big(\log(1+\sin x)-\log(1+\sin x)\Big)\]
*\[\frac{1}{2}\Big(\log(1+\sin x)-\log(1-\sin x)\Big)\]
Oh yeah, I need absolute values. \[\frac{1}{2}\Big(\log|1+\sin x|−\log|1−\sin x|\Big)\]Which is also\[\frac{1}{2}\log\frac{|1+\sin x|}{|1−\sin x|}\], but now I'm just cheating off Wikipedia.
@MathSofiya still want help with\[\int\cos x\sec^2xdx\]?
it seems that everyone here has forgotten the simple trick to this integral...
see there had to be a simpler way
\[\cos x\sec^2x=\cos x\cdot\frac1{\cos^2 x}=\frac1{\cos x}=\sec x\]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)
\[\int\sec x\left({\sec x+\tan x\over\sec x+\tan x}\right)dx=\int{\sec^2x+\sec x\tan x\over\sec x+\tan x}dx\]magically the top is now the derivative of the bottom:\[u=\sec x+\tan x\implies du=\sec^2x+\sec x\tan xdx\]\[\int\frac1udu=\ln|u|+C=\ln|\sec x+\tan x|+C\]
that is the standard approach for this well-known integral
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.
but yes it makes sense now I see it.
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 non-obvious 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 :)
I shall memorize this integral =P (honestly I will) And as always, Thanks Turing!

Not the answer you are looking for?

Search for more explanations.

Ask your own question