Why Does Integrating 2/x Yield Different Results?

ryan8888
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Hi all,

I am trying to work out the following:

cos2(t)sin(t)y' = -cos3(t)y + 1

I've moved put the equation into the standard form and determined that my integrating factor is the integral of cos(t)/sin(t) or the integral of cot(t). This gives me elnsinu +c or sin(t).

I multiple both sides and simplify and I get to this point"

y*sin(t) = Integral of: 1/cos2(t) and I'm having a hard time finding the integral of this function.

Once I solve the integral on the right hand side I'm home free but can't figure it out.

Any help is appreciated.

Thanks

Ryan
 
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\int \frac{dt}{cos^2(t)} = \int sec^2(t)dt = tan(t) + C

Caveat: I didn't verify the work leading up to your question.
 
This is just what I was looking for. I didn't have that particular integral in my list of variations, do you happen to have a list of some of these less seen integrals of trig functions?

Thanks for all of your help!
 
No, I don't have a list. This is a very common formula that you should have committed to memory. It comes directly from the derivative formulas for the trig functions.

d/dx(sin x) = cos x
d/dx(cos x) = -sin x
d/dx(tan x) = sec2 x
d/dx(sec x) = sec x * tan x
d/dx(cot x) = -csc2 x
d/dx(csc x) = -csc x * cot x

The integrals of each expression on the right is the corresponding expression on the left without the differentiation operator. For example, \int sec x tan x dx = sec x + C, and so on for all the others.
 
Thanks again Mark. I just havn't come across that one to this point! But it is very helpful. I have one more unrelated one for you. I have to integrate 2/x. Now I come up with 2 * ln(x) but in my textbook example it shows the solution as ex2, so the answer would be x2. Am I missing something simple with this integral?


Mark44 said:
No, I don't have a list. This is a very common formula that you should have committed to memory. It comes directly from the derivative formulas for the trig functions.

d/dx(sin x) = cos x
d/dx(cos x) = -sin x
d/dx(tan x) = sec2 x
d/dx(sec x) = sec x * tan x
d/dx(cot x) = -csc2 x
d/dx(csc x) = -csc x * cot x

The integrals of each expression on the right is the corresponding expression on the left without the differentiation operator. For example, \int sec x tan x dx = sec x + C, and so on for all the others.
 
Never mind! I just realized that ln x^r = r ln x!

That answers that question!

Thanks for you help

Ryan
 
ryan8888 said:
Thanks again Mark. I just havn't come across that one to this point! But it is very helpful. I have one more unrelated one for you. I have to integrate 2/x. Now I come up with 2 * ln(x) but in my textbook example it shows the solution as ex2, so the answer would be x2. Am I missing something simple with this integral?
\int \frac{2}{x} = 2 ln|x| + C

If they came up with ex2, I don't have any idea what they did, so you would need to show me the example.
 
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