| Thread Closed |
Help with IntegralS! Very Important! Quick! |
Share Thread |
| May6-07, 06:09 AM | #35 |
|
|
Help with IntegralS! Very Important! Quick!
I start do 6th and.. have problems in middle. see in atach
|
| 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 |
|
Recognitions:
|
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 |
|
Recognitions:
|
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 |
|
Recognitions:
|
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?
|
| May6-07, 06:47 AM | #42 |
|
Recognitions:
|
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 |
|
|
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.
|
| 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 =*(
|
| Thread Closed |
Similar discussions for: Help with IntegralS! Very Important! Quick!
|
||||
| Thread | Forum | Replies | ||
| quick question about multiple integrals | General Math | 5 | ||
| A Few Quick Questions: Integral Notation and Integrals | Calculus & Beyond Homework | 4 | ||
| I need help quick! Integrals | Calculus & Beyond Homework | 3 | ||