Question

\[\int cosxsec^2xdx\]
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?

Answer 1

Asking cos*sec^2, not cos/sec^2

Answer 2

Oh, sorry:
Cosx/Cos^2x = 1/Cosx = Secx

Answer 3

Well actually you'd keep it as 1/Cox, and use integration by parts

Answer 4

wait a second
\[\int cosx\frac 1{cos^2x}dx?\]

Answer 5

Yes.

Answer 6

Sorry for the confusion.. Let's start over:
Cosx*Sec^2x = Cosx/Cos^2x = 1/Cosx, since Secx = 1/Cosx

Answer 7

\[\int\frac 1{cosx}dx\]

Answer 8

\[\int{\cos{x}*\frac{1}{1-\sin^2{x}}}\ dx\]

Answer 9

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

Answer 10

why did you change it? wouldn't it make more sense to cancel one cosine out?

Answer 11

No

Answer 12

Let u = sin(x)
du = cos(x)dx
\[\int{\frac{du}{1-u^2}}\]

Answer 13

Actually this would be the easiest option, imo:
Cos^2(u) = (1/2)+(Cos(2u))/2

Answer 14

ok let Valpey, finish this, otherwise you'll get confused.

Answer 15

sure

Answer 16

do I multiply the integral by \[\ frac {1+u^2}{1+u^2}\]

Answer 17

\[\frac {1+u^2}{1+u^2}\]

Answer 18

\[ \int \sec x dx = \log \left | \sec x + \tan x\right | = \log \left| \tan \left( {x \over 2} + {\pi \over 4}\right)\right | \]
wikipedia says it's quite a problem
http://en.wikipedia.org/wiki/Integral_of_the_secant_function

Answer 19

makes sense, thank you!

Answer 20

Oh, duh! My bad.\[\int\frac{1}{1-u^2}du=\int\frac{1}{(1-u)(1+u)}du=\frac{1}{2}\int\frac{1+1}{(1-u)(1+u)}du\]\[=\frac{1}{2}\int\Big(\frac{1+u}{(1-u)(1+u)}+\frac{1-u}{(1-u)(1+u)}\Big)du\]\[=\frac{1}{2}\int\frac{1+u}{(1-u)(1+u)}du+\frac{1}{2}\int\frac{1-u}{(1-u)(1+u)}du=\frac{1}{2}\int\frac{1}{1-u}du+\frac{1}{2}\int\frac{1}{1+u}du\]
\[=\frac{1}{2}(-\log(1-u))+\frac{1}{2}(\log(1+u))\]
\[\frac{1}{2}\Big(\log(1+\sin x)-\log(1+\sin x)\Big)\]

Answer 21

*\[\frac{1}{2}\Big(\log(1+\sin x)-\log(1-\sin x)\Big)\]

Answer 22

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.

Answer 23

@MathSofiya still want help with\[\int\cos x\sec^2xdx\]?

Answer 24

it seems that everyone here has forgotten the simple trick to this integral...

Answer 25

see there had to be a simpler way

Answer 26

\[\cos x\sec^2x=\cos x\cdot\frac1{\cos^2 x}=\frac1{\cos x}=\sec x\]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)

Answer 27

\[\int\sec x\left({\sec x+\tan x\over\sec x+\tan x}\right)dx=\int{\sec^2x+\sec x\tan x\over\sec x+\tan x}dx\]magically the top is now the derivative of the bottom:\[u=\sec x+\tan x\implies du=\sec^2x+\sec x\tan xdx\]\[\int\frac1udu=\ln|u|+C=\ln|\sec x+\tan x|+C\]

Answer 28

that is the standard approach for this well-known integral

Answer 29

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.

Answer 30

but yes it makes sense now I see it.

Answer 31

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 :)

Answer 32

I shall memorize this integral =P (honestly I will)
And as always, Thanks Turing!