# Integrating: 30 Years Later - No Taylor Book

• denverdoc
In summary, the conversation was about integrating the expression Int(dv/(k-v^2))=dt, with the speaker mentioning a substitution of v=k1/2sin(x). They also discussed the different cases for k and provided hints for solving the integral. Finally, the speaker asked for recommendations for a good calculus textbook.

#### denverdoc

for the life of me I cannot recall how to integrate say

Int(dv/(k-v^2))=dt

Its been 30 years since I took calc, and can't find my Taylor book.

Try the substitution v=k1/2sin(x)

denverdoc said:
for the life of me I cannot recall how to integrate say

Int(dv/(k-v^2))=dt

Its been 30 years since I took calc, and can't find my Taylor book.
Why is there a dt thingy following the integral expression? So, I assume that you want to integrate:
$$\int \frac{dv}{k - v ^ 2}$$
If k > 0, you can do something like this:
$$\int \frac{dv}{k - v ^ 2} = \int \frac{dv}{(\sqrt{k} - v) (\sqrt{k} + v)} = \frac{1}{2 \sqrt{k}} \int \frac{(\sqrt{k} - v) + (\sqrt{k} + v)}{(\sqrt{k} - v) (\sqrt{k} + v)} dv$$
$$= \frac{1}{2 \sqrt{k}} \int \left( \frac{1}{\sqrt{k} + v} + \frac{1}{\sqrt{k} - v} \right) dv = ...$$
Can you go from here? :)
If k < 0 (or in other words, -k > 0), then you can pull out the minus sign, like this:
$$\int \frac{dv}{k - v ^ 2} = - \int \frac{dv}{- k + v ^ 2} = - \int \frac{dv}{(\sqrt{- k}) ^ 2 + v ^ 2} = -\frac{1}{\sqrt{-k}} \arctan \left( \frac{x}{\sqrt{-k}} \right) + C$$, can you get this? :)

VietDao29 said:
Why is there a dt thingy following the integral expression? So, I assume that you want to integrate:
$$\int \frac{dv}{k - v ^ 2}$$
If k > 0, you can do something like this:
$$\int \frac{dv}{k - v ^ 2} = \int \frac{dv}{(\sqrt{k} - v) (\sqrt{k} + v)} = \frac{1}{2 \sqrt{k}} \int \frac{(\sqrt{k} - v) + (\sqrt{k} + v)}{(\sqrt{k} - v) (\sqrt{k} + v)} dv$$
$$= \frac{1}{2 \sqrt{k}} \int \left( \frac{1}{\sqrt{k} + v} + \frac{1}{\sqrt{k} - v} \right) dv = ...$$
Can you go from here? :)
If k < 0 (or in other words, -k > 0), then you can pull out the minus sign, like this:
$$\int \frac{dv}{k - v ^ 2} = - \int \frac{dv}{- k + v ^ 2} = - \int \frac{dv}{(\sqrt{- k}) ^ 2 + v ^ 2} = -\frac{1}{\sqrt{-k}} \arctan \left( \frac{x}{\sqrt{-k}} \right) + C$$, can you get this? :)

sure--er, I think... in the first case, let u=(sqrt(k)+v)
du=dv then becomes, int(du/u)... =ln(sqrt(k)+v)-ln(sqrt(k)-v)+C=

using deletes hint, v=k^1/2sin:
k^-1/2int(cos(x)/(k(1-sin^2(x)) which can be treated similarly or simply looked up as the integral of secant.

Thanks so much, the memories are coming back.

Can anyone recommend a good calculus text? I still have my differential eqns/engineering math text, just need a good basic text..

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Plz, help :

int_{- infinity}^{+ infinity} dk [exp(ikx)]/(k^2+a^2) -?

Lol mate you can't just find someone else's thread about an integral and put yours in it.