definite integrals

[Leaping antalope]

can anyone help me with this problem...
the integral from 0 to half pi: 1/(1+cosx)dx...
Thanks...

[arildno]

Hint use the relation:
\cos^{2}(\frac{x}{2})=\frac{1+\cos(x)}{2}

[teclo]

from my perspective, that problem would be very hard to solve without changing it somehow. try multiplying the top and bottom by (1-cos x) and see what happens. remember that sin^2 x + cos^2 x = 1.

that is the method i used, but it did require me to use the reduction formula on one of the integrals. hopefully that helps, thought it may not be the inteded method.

[Leaping antalope]

so 1/(1+cosx)=(1-cosx)/(1-cos^2x)=(1-cosx)/sin^2x=(1/sin^2x)-(cosx/sin^2x)...
Aha~now i get it
the integration of 1/sin^2x is -cotx, and cosx/sin^2x=cosx*sin^(-2)x, and the integral of this is -1/sinx
so the integration of 1/(1+cosx) is -cotx+1/sinx, but now i have a problem, cot(half pi) does not exist...
help still needed...

[teclo]

so 1/(1+cosx)=(1-cosx)/(1-cos^2x)=(1-cosx)/sin^2x=(1/sin^2x)-(cosx/sin^2x)...
Aha~now i get it
the integration of 1/sin^2x is -cotx, and cosx/sin^2x=cosx*sin^(-2)x, and the integral of this is -1/sinx
so the integration of 1/(1+cosx) is -cotx+1/sinx, but now i have a problem, cot(half pi) does not exist...
help still needed...

i forgot that the derivative of cot is -csc^2. you don't need the reduction formula after all! i worked this out and i got

1/sin x - cot x
or
1/sin x - cos x/sin x

the problem is when it approaches zero, not pi/2. to me it looks like this is divergent, thus there is no definant area (as it approaches zero from the right, 1/sin x approaches infinity).

maybe i did something wrong, but the math looks correct. as a forewarning i'm just finishing the last chapter in calc 2 in college, so i'm not a math guru.

edit: i don't remember if limit laws dictate that you could say infinity minus infinity equals a total of zero. that could be it's been almost a year since i took calc 1. if so, you'd have (1-0)-(infinity - infinity = 0) = 1 unit^2.

i don't know when i get problems wrong on tests its usually due to magic algebra, that could be a case of it right there. i'd look it up but i need to finish writing up my lab for this evening.

[Leaping antalope]

The answer is 1. Or maybe we should convert (1/sin^2x)-(cosx/sin^2x) so that we can substitute half pi and 0 in it...not sure...

[teclo]

The answer is 1. Or maybe we should convert (1/sin^2x)-(cosx/sin^2x) so that we can substitute half pi and 0 in it...not sure...

well, sin of 0 is 0, so either way you're dividing both by zero. the only thing i can figure out is that you do it as an improper integral and the infinities cancel eachother.

[marlon]

Hint use the relation:
\cos^{2}(\frac{x}{2})=\frac{1+\cos(x)}{2}

I suggest you use the advice of arildno. It will be the best way to solve this.

\int\frac{2}{\cos(x)+1}*dx = \int\frac{1}{\cos^{2}(\frac{x}{2})} * dx

and you know that dx = 2*d(\frac{x}{2})
and
\int\frac{1}{\cos^{2}(x)} = \tan(x)

regards
marlon

[teclo]

I suggest you use the advice of arildno. It will be the best way to solve this.

\int\frac{2}{\cos(x)+1}*dx = \int\frac{1}{\cos^{2}(\frac{x}{2})} * dx

and you know that dx = 2*d(\frac{x}{2})
and
\int\frac{1}{\cos^{2}(x)} = \tan(x)

regards
marlon

that is clever, i didn't remember that identity. but you aren't finished. remember that the integral is from 0 to pi/2. under that integral the function is divergent, thus the area doesn't exist. right? (lim t---> pi/2 [tan x] from 0 to t) sin x /cos x -- as x approaches pi/2, thus cos x = 0, bad things happen.

also divergent by relation to 1/cos x isn't it?

if anyone has a certain answer, please elucidate! (no mathmatica at home :( )

[teclo]

I suggest you use the advice of arildno. It will be the best way to solve this.

\int\frac{2}{\cos(x)+1}*dx = \int\frac{1}{\cos^{2}(\frac{x}{2})} * dx

and you know that dx = 2*d(\frac{x}{2})
and
\int\frac{1}{\cos^{2}(x)} = \tan(x)

regards
marlon

edit: nevermind, i thought you were saying the answer was tan x. disregard this and above message!