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Old 06-14-2007   #1
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Default Integral and exponentials

I'm working on a 19 page assignment for my calculus course. Unfortunately this is a distance course, and there aren't many places I can go to for help... so bre with me.

Question #1

(integral) (2 + 3cosx + (sinx)^3) / ((sinx)^2) dx

Developing... ((integral) (2/(sinx)^2) + (3cosx)/((sinx)^2) + sinx) dx

Might want to write that out before attempting since it's a little difficult to read it properly when typed out.

Note that (sinx)^3 should usually be written out (sin^3)x.

Question #2

(1/4)^x * 16^(x/2)

The answer, I believe, is 1. However I can't seem to reach this by applying exponential laws.

Last edited by Mr. Hui; 06-14-2007 at 05:04 PM..
Old 06-15-2007   #2
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Ok I worked at the first problem and figured it out.

using the identity 1/sinx = cscx...

(2/(sinx)^2)x = 2(csc^2)x

Using the identity cosx/sinx = cotx

3cosx/((sinx)^2) = 3cotx/sinx

And using the identity 1/sinx = cscx

3cotx/sinx = 3cscxcotx

Giving a final developed equation of
(integral) (2(csc^2)x + 3cscxcotx + sinx) dx

Remembering these derivatives
cosx = -sinx
cscx = -cscxcotx
tanx = (csc^2)x

Final integrated answer = -2cotx - 3cscx - cosx + C

C being any integer.

But I still can't figure out the second one, although it's probably much easier
Old 06-15-2007   #3
Mr. Hui

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Is the second problem an equation?
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Old 06-15-2007   #4
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Sorry, yes the 2nd was an equation. I played around with it and seem to have figured it out as well.

f(x) = (1/4)^x * 16^(x/2)

(1^x / 4^x) * 16^(x/2)

1^x = 1 so

f(x) = 1/(4^x) * 16^(x/2)
= 16^(x/2) / (4^x)
= (sqrt16)^x / (4^x)
= 4^x / 4^x
= 1

I know I know... ended up figuring it by myself. However I did end up wasting hours while trying to figure it out.

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