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Raithed
China7078 Posts
This is my takehome test that is due tomorrow. Unfortunately, I do not know how to do any of these... Explanations on how to integrals would be greatly appreciated. I am working on these on my own, and would like ideas, or someone to talk to on AIM or something to discuss how to do these. ![[image loading]](http://img134.imageshack.us/img134/2503/img001zq3.jpg)  | ||
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fight_or_flight
United States3988 Posts
![[image loading]](http://img233.imageshack.us/img233/3386/reviewzu8.png) | ||
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fight_or_flight
United States3988 Posts
![[image loading]](http://img138.imageshack.us/img138/483/review2fr3.png) | ||
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Raithed
China7078 Posts
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fight_or_flight
United States3988 Posts
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doob10163
6 Posts
I have 1, 2, 3, 4, 6, 10, and 14 done Not quite sure how to go about the rest of these. Any ideas? | ||
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fight_or_flight
United States3988 Posts
cos^3 = cos cos^2 = cos (1 - sin^2) then u = sin, du = 1/cos dx. For the rational ones do partial fractions. edit: also, you can use http://integrals.wolfram.com/index.jsp to check your work | ||
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doob10163
6 Posts
i did 4 times the integral (sec^2 x) ^2 which then became sex^2 x (tan ^2x + 1) is this the right approach to be taking? If so , what is the proper next step? | ||
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paper
13196 Posts
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Saracen
United States5139 Posts
for 9. "which became sex^2x" ahah i see what you did there ^_^ sec^2(x)tan^2(x) function to a power times its derivative and S sec^2(x) dx = tanx well at least i'm pretty sure how it is i forgot all this shitty trig identity integration stuff | ||
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fight_or_flight
United States3988 Posts
then its the integral of u^2 + 1 = 1/3 u^3 + u or 1/3 (tanx)^3 + tanx giving 4/3 (tanx)^3 + 4tanx | ||
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Saracen
United States5139 Posts
right?  | ||
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doob10163
6 Posts
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fight_or_flight
United States3988 Posts
Basically I've just been working off my own flowchart to be honest. | ||
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doob10163
6 Posts
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doob10163
6 Posts
The second integral of #3 being what happens after you split the fraction and are left with 18x over the denominator | ||
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fight_or_flight
United States3988 Posts
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Saracen
United States5139 Posts
On April 02 2008 12:55 doob10163 wrote: Any ideas on how to do the 2nd integral of 3, 5, 11, 12, 13, 15, 16, 17, 18, and 19? The second integral of #3 being what happens after you split the fraction and are left with 18x over the denominator separate it into 2 fractions 1/(1+3x^2) and -6x/(1+3x^2) after you take out the 3 1st one is arctan 2nd one is function to a power times its derivative | ||
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Saracen
United States5139 Posts
u-sub u=lnx du=dx/x x=e^u then use parts (tabular method) | ||
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Saracen
United States5139 Posts
it's really straightforward...idk why it's hard? sec(theta) = sqrt(4y^2+1), dy=(sec^2(theta))d(theta)/2 | ||
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goldrush
Canada709 Posts
A/(x+2) + B/(x-2) + C/x = (x+4)/(x^3-4x), solve for each one and then take the integral of the individual fractions. That's the partial fraction method. | ||
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Saracen
United States5139 Posts
the derivative of -x^2+4x+5 = ?!? -2x + 4 which just happens to be -2 times (x-2) oh snap function to a power times derivative | ||
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doob10163
6 Posts
On April 02 2008 13:30 Saracen wrote: 11 is just trig sub... it's really straightforward...idk why it's hard? sec(theta) = sqrt(4y^2+1), dy=(sec^2(theta))d(theta)/2 What exactly am I substituting for what here? Thanks for the help, by the way, i really appreciate it. | ||
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goldrush
Canada709 Posts
Since you're not familiar with this stuff, don't try and take shortcuts. Use #17 as an example: Construct a right-angle triangle with a hypotenuse at sqrt2 and the two other sides as x and sqrt(2-x^2). Using pythagoras, you can tell it's right (x^2 + 2 - x^2 = 2, which is the square of the hypotenuse). Alright now... using this triangle, deduce that sin theta = x / sqrt2. Therefore, sqrt 2 * sin theta = x, sqrt 2 cos theta dtheta = dx. You still need to replace sqrt(2-x^2) and x^2 in the original equation though, so... cos theta = sqrt (2- x^2) / sqrt2 sqrt (2 - x^2) = sqrt2 * cos theta original equation of sin theta = x / sqrt2, deduce that x^2 = 2 sin^2 theta. Now, plug it all into the formula: sqrt 2 costheta dtheta / (2 sin^2 theta * sqrt 2 * cos theta) Simplifies oput to 0.5 dtheta/ sin^2 theta. Integrating this turns out to be -0.5 costheta / sintheta. But this is a definite integral and the bounds are also changed becuase of the variable change. Using sintheta = x / sqrt2, plug in 1 and sqrt 2 in the place of x and solve for theta. It turns out to be theta = 90 and 45 degrees. So plug it in... -0.5 [ cos 90 / sin 90 - cos 45 / sin 45] = -0.5 [ 0 / 1 - 1 ] = 0.5 Apologies for any mistakes, it's really late over here. hope this helped.  | ||
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