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Please help me with this integration
- Thread starter oguzkan
- Start date
Is this your problem:
\(\displaystyle \int\frac{1}{\sqrt{u(au-b)^{2}-4u}}du=\int\frac{1}{\sqrt{u[(au-b)^{2}-4]}}du\)
You did not say what you are integrating with respect to. Is it u?. I would presume so,
Now, if we let \(\displaystyle t=au-b, \;\ \frac{dt}{a}=du, \;\ u=\frac{t+b}{a}\), it may at first look promising. But that extra u in there makes it a little rougher.
We get \(\displaystyle \frac{1}{a}\int\frac{1}{\sqrt{\frac{(t+b)(t^{2}-4)}{a}}}dt\)
If that is it, it is rather tough and may involve Elliptic Integrals. Maple gave me a long, drawn out, convoluted solution involving Elliptic integrals.
Therefore, to do it by hand may prove daunting. May I ask what class this is from and at what level you are?.
Some of the other skilled mathematicians may have a good idea. I do not have time right now, but if you do not get a response, I can look later this afternoon.
\(\displaystyle \int\frac{1}{\sqrt{u(au-b)^{2}-4u}}du=\int\frac{1}{\sqrt{u[(au-b)^{2}-4]}}du\)
You did not say what you are integrating with respect to. Is it u?. I would presume so,
Now, if we let \(\displaystyle t=au-b, \;\ \frac{dt}{a}=du, \;\ u=\frac{t+b}{a}\), it may at first look promising. But that extra u in there makes it a little rougher.
We get \(\displaystyle \frac{1}{a}\int\frac{1}{\sqrt{\frac{(t+b)(t^{2}-4)}{a}}}dt\)
If that is it, it is rather tough and may involve Elliptic Integrals. Maple gave me a long, drawn out, convoluted solution involving Elliptic integrals.
Therefore, to do it by hand may prove daunting. May I ask what class this is from and at what level you are?.
Some of the other skilled mathematicians may have a good idea. I do not have time right now, but if you do not get a response, I can look later this afternoon.
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
\(\displaystyle Letting \ t = au-b \ and \ then \ letting \ t = 2sin(\theta), \ I \ get:\)
\(\displaystyle \frac{-1}{a^{1/2}}\int\frac{d\theta}{[2sin(\theta)+b]^{1/2}}\)
\(\displaystyle Now, \ I \ can't \ go \ any \ further.\)
\(\displaystyle \frac{-1}{a^{1/2}}\int\frac{d\theta}{[2sin(\theta)+b]^{1/2}}\)
\(\displaystyle Now, \ I \ can't \ go \ any \ further.\)
Let's do it this way:
\(\displaystyle \int\frac{1}{\sqrt{u(au-b)^{2}-4u}}du\)
\(\displaystyle =\frac{-1}{a\sqrt{u(au-b+2)(au-b-2)}}\left[2(b+2)\sqrt{\frac{a(u-\frac{b+2}{a})}{b+2}}\sqrt{\frac{u-\frac{b-2}{a}}{\frac{b+2}{a}-\frac{b-2}{a}}}\sqrt{\frac{au}{b+2}}\cdot \text{EllipticF}\left(\sqrt{\frac{a(\frac{b+2}{a}-u)}{b+2}}, \;\ \sqrt{\frac{b+2}{a(\frac{b+2}{a}-\frac{b-2}{a})}}\right)\right]\)
That is what Maple 10 gave me. When I said I thought it may involve an Elliptic Integral, apparently I was correct to some extent.
Now, if this thing had some limits of integration, then we may be able to do something. But, the indefinite form is a booger.
As I stated before, you did not say what we were integrating with respect to. If we integrate with respect to a or b, then we can do it easier and get a form. But u is a monster
\(\displaystyle \int\frac{1}{\sqrt{u(au-b)^{2}-4u}}du\)
\(\displaystyle =\frac{-1}{a\sqrt{u(au-b+2)(au-b-2)}}\left[2(b+2)\sqrt{\frac{a(u-\frac{b+2}{a})}{b+2}}\sqrt{\frac{u-\frac{b-2}{a}}{\frac{b+2}{a}-\frac{b-2}{a}}}\sqrt{\frac{au}{b+2}}\cdot \text{EllipticF}\left(\sqrt{\frac{a(\frac{b+2}{a}-u)}{b+2}}, \;\ \sqrt{\frac{b+2}{a(\frac{b+2}{a}-\frac{b-2}{a})}}\right)\right]\)
That is what Maple 10 gave me. When I said I thought it may involve an Elliptic Integral, apparently I was correct to some extent.
Now, if this thing had some limits of integration, then we may be able to do something. But, the indefinite form is a booger.
As I stated before, you did not say what we were integrating with respect to. If we integrate with respect to a or b, then we can do it easier and get a form. But u is a monster