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Help with IntegralS! Very Important! Quick!

 
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May6-07, 06:09 AM   #35
 

Help with IntegralS! Very Important! Quick!


I start do 6th and.. have problems in middle. see in atach
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May6-07, 06:15 AM   #36
 
so the rsult of 5: [tex]\intdu=\arcsin \frac{x}{2} + C[/tex] ?
May6-07, 06:15 AM   #37
 
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I have no idea what the "tg" is, but I deduce its the tangent function.

Even if it isnt, let [tex]u=\sqrt{x}[/tex] Then [tex]\frac{du}{dx} = \frac{1}{2u}, dx = 2u du[/tex]

so the integral becomes

[tex]\int \tg u \frac{2u}{u} du = 2\int \tg (u) du[/tex]

EDIT: As to post 36, q5, yes that is the answer. The reason you can not write it as you did previously is because it is only valid for n E Z, but we need it to be valid for all real values of n.
May6-07, 06:20 AM   #38
 
Gibs [tex]u=\sqrt{x}[/tex]
[tex]du=\frac{1}{2\sqrt{x}}dx[/tex] then [tex]dx=2\sqrt{x}du[/tex]
May6-07, 06:23 AM   #39
 
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Check your typing, you forgot to use the right hand brace } instead of right breacket ). Anyway, then you are correct. However, didn't we say [itex]u=\sqrt{x}[/itex]?
May6-07, 06:24 AM   #40
 
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You know either way it doesn't matter. You are stuck with
[tex]2\int \frac{\sin x}{\cos x} dx[/tex] which can be solved by letting u = cos x
May6-07, 06:40 AM   #41
 
right result?
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May6-07, 06:47 AM   #42
 
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Nope.

[tex]2\int \frac{\sin x}{\cos x} dx[/tex]
let u = cos x, then du = - sin x dx
[tex] -2\int \frac{1}{u} du = -2\log_e u + C = -2 \log_e (\cos x) + C[/tex]
May6-07, 07:13 AM   #43
 
Quote by Gib Z View Post
Nope.

[tex]2\int \frac{\sin x}{\cos x} dx[/tex]
let u = cos x, then du = - sin x dx
[tex] -2\int \frac{1}{u} du = -2\log_e u + C = -2 \log_e (\cos x) + C[/tex]
U right, i agree with u. I made a child mistake =(
simply I sit near computer and do homework already 6 hours and have square head
May6-07, 07:15 AM   #44
 
start the 7th. Stop in the middle. See in atach.
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May6-07, 07:26 AM   #45
 
Mentor
So you have the integral of sec u= 1/cos u. To compute this, try multiplying top and bottom by sec u+tan u.
May6-07, 07:32 AM   #46
 
i don`t understand u cristo =(
May6-07, 07:39 AM   #47
 
stop don`t tell me anything 5 minut i try tio solve it again.
May6-07, 07:45 AM   #48
 
secu^2= 1+tgu^2 => secu=sqrt(1+tgu^2)
so compare 3rd step and last in atach . hmm
May6-07, 07:46 AM   #49
 
Mentor
[tex]\sec u=\sec u\cdot\left(\frac{\sec u+\tan u}{\sec u+\tan u}\right)[/tex]. Can you integrate this?

[Hint: How is the numerator related to the denominator?]
May6-07, 07:48 AM   #50
 
ou... sorry =) how to delete.... -) it`s mistake
May6-07, 07:50 AM   #51
 
to cristo... no i can`t =*(
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