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.

fight_or_flight

United States3988 Posts

April 02 2008 02:05 GMT

fight_or_flight

United States3988 Posts

April 02 2008 02:17 GMT

Raithed

China7078 Posts

April 02 2008 02:20 GMT

ROFL. Those are EPIC.

fight_or_flight

United States3988 Posts

April 02 2008 02:22 GMT

That last one is exclusive content btw (made it myself).

doob10163

6 Posts

April 02 2008 02:28 GMT

Thanks guys. This is the friend that Raithed posted for

I have 1, 2, 3, 4, 6, 10, and 14 done

Not quite sure how to go about the rest of these. Any ideas?

fight_or_flight

United States3988 Posts

April 02 2008 02:40 GMT

The flowchart should help with some of them. Remember, sin^2 = 1 - cos^2 and cos^2 = 1 - sin^2. That means if a power is odd you can

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

doob10163

6 Posts

April 02 2008 02:50 GMT

My trig is terrible - for example, I would have no idea how to get about #9.

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?

paper

13197 Posts

April 02 2008 02:53 GMT

damn i wish i had that shit fight/flight posted when i took calc :{

Saracen

United States5139 Posts

April 02 2008 03:10 GMT

#10

you can do most of these by parts or trig identities

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

fight_or_flight

United States3988 Posts

April 02 2008 03:12 GMT

#11

so then you have u = tan, then the derivative of u, du, becomes dx = 1/sec^2 du.

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

Saracen

United States5139 Posts

April 02 2008 03:16 GMT

#12

you don't need to use u-substitution on that one
right?

doob10163

6 Posts

April 02 2008 03:24 GMT

#13

Thanks, fight_or_flight. Do you have any idea on how to go about #5? (Do you have AIM, MSN, or YIM and don't mind me messaging you for help?)

fight_or_flight

United States3988 Posts

April 02 2008 03:35 GMT

#14

actually, I should be doing other things besides this, and I have to go home soon. Anyway, I haven't taken calc in a long time so these problems are hard for me too, lol.

Basically I've just been working off my own flowchart to be honest.

doob10163

6 Posts

April 02 2008 03:35 GMT

#15

Thanks anyways

doob10163

6 Posts

April 02 2008 03:55 GMT

#16

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

fight_or_flight

United States3988 Posts

April 02 2008 03:58 GMT

#17

u = 1+3x^2, du = 6xdx?

Saracen

United States5139 Posts

April 02 2008 04:20 GMT

#18

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

Saracen

United States5139 Posts

April 02 2008 04:25 GMT

#19

number 5:
u-sub
u=lnx
du=dx/x
x=e^u
then use parts (tabular method)

Saracen

United States5139 Posts

April 02 2008 04:30 GMT

#20

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

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