Integral of secant

  • Thread starter Sisyphus
  • Start date
  • #1
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Homework Statement



Basically, I have to find

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

by multiplying the integrand by [tex] \frac{cosx}{cosx}


[/tex]

I go through and arrive at a solution, but when I differentiate it,
I get -tan(x)

something's clearly wrong, but I can't see what it is that I'm doing wrong here...


Homework Equations



[tex]
let u = sin(x)[/tex]


[tex] du = cos(x)dx
[/tex]

The Attempt at a Solution



[tex]
\int \frac{1}{cosx} dx = \int \frac{cosx}{cos^2x} dx\\
= \int \frac{cosx}{1-sin^2x} dx\\
=\int \frac{du}{1-u^2} \\
=\int \frac{du}{(1-u)*(1+u)} \\
=\frac{1}{2} * \int \frac {1}{1+u} + \frac {1}{1-u} du\\
= \frac{1}{2} * (ln(1-u^2}})
=\frac{1}{2} * (ln(cos^2))


[/tex]
 
Last edited:

Answers and Replies

  • #2
StatusX
Homework Helper
2,564
1
Your second to last step (where you actually perform the integration) is wrong. In the second term, remember that u has a minus sign.
 
  • #3
61
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hmm..

I've never run into anything like this before

so why does u having a minus sign in front of it pose a problem with what I did in my original solution?

(thanks for the help)
 
  • #4
cristo
Staff Emeritus
Science Advisor
8,107
73
Well, the value of your integral will be ln(1+u)-ln(1-u)
 
  • #5
61
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if I kept the minus sign where it was, which is what I did in my original solution, my integral would've been
[tex] ln(1+u)+ln(1-u)=ln(1-u^2)[/tex]

StatusX told me to watch out for the negative sign in front of the u, so I factored it out, made sure that u was positive, and then integrated it, which gave me the correct solution.

the thing is i'm not sure why I couldn't proceed as usual with the minus sign in front of the u
 
  • #6
cristo
Staff Emeritus
Science Advisor
8,107
73
StatusX said to note the minus sign in front of the u. You can proceed as normal, but noting that [tex]\int\frac{1}{1-u}du=-ln(1-u)[/tex]. In general [tex]\int\frac{1}{f(u)}du=\frac{ln[f(u)]}{df/du}[/tex]. In this case, f(u)=1-u, and so df/du=-1
 
  • #7
61
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ah, ok

thank you!
 
  • #8
o_O i didn't know we would utilizie such a method to do this integral.

i've always thought the integral of secant was just sec[x]tan[x]!
 
  • #9
cristo
Staff Emeritus
Science Advisor
8,107
73
o_O i didn't know we would utilizie such a method to do this integral.

i've always thought the integral of secant was just sec[x]tan[x]!
That's the derivative of sec(x)
 

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