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

goldrush

Canada709 Posts

April 02 2008 04:33 GMT

#21

If you know how to do partial fractions, then questions like #15 are done by first taking out the 1/3 (it's a lot easier), then factoring out the denominator into x+2, x-2 and x. You assign a variable such as A/B/C to each numerator. Therefore:

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.

Saracen

United States5139 Posts

April 02 2008 04:33 GMT

#22

12. is also easy...
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

doob10163

6 Posts

April 02 2008 05:00 GMT

#23

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.

goldrush

Canada709 Posts

April 02 2008 06:42 GMT

#24

Well, I'll try and help you a tad before I go to bed.

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.