Integrating ln(sqrt(t)/t) using u-substitution

[judahs_lion]

Homework Statement

integrate lnsqrt(t)/t , dt

Homework Equations

The Attempt at a Solution

This is the most i can come up with.

u = lnsqrt(t), so du = 1/(2sqrt(t))dt , so dt = 2du/sqrt(t)

Im stuck @ this point

 

[╔(σ_σ)╝]

Hint: The logarithm function has an interesting property.

$$ln\left(a^{b} \right) = bln\left(a \right)$$

 

[Dick]

du isn't 1/(2*sqrt(t))*dt. You are forgetting to use the chain rule. You can also get the correct differential by applying the rules of logs to ln(sqrt(t))=ln(t^(1/2)).

 

[judahs_lion]

Hint: The logarithm function has an interesting property.

$$ln\left(a^{b} \right) = bln\left(a \right)$$

so i would integrate (lnt^(1/2))/t ?

 

[Dick]

so i would integrate (lnt^(1/2))/t ?

Sure. Isn't ln(sqrt(t))/t=ln(t^(1/2))/t?

 

[judahs_lion]

Sure. Isn't ln(sqrt(t))/t=ln(t^(1/2))/t?

Ok, now I'm not sure what to take as u.

 

[╔(σ_σ)╝]

so i would integrate (lnt^(1/2))/t ?

Yes.
But follow Dicks' suggestion also. It would be helpful to you. Your original method was correct ,however, as already pointed out you weren't differentiating correctly.

du isn't 1/(2*sqrt(t))*dt. You are forgetting to use the chain rule. You can also get the correct differential by applying the rules of logs to ln(sqrt(t))=ln(t^(1/2)).

 

[Dick]

Ok, now I'm not sure what to take as u.

? Take u the same as you originally chose. u=ln(sqrt(t))=ln(t^(1/2)). They are the same thing. You just have to compute du correctly. That's where your first attempt went wrong. The du was wrong.

 

[judahs_lion]

Yea i think i integrated rather then differentiated to get du. so u = 1/2lnt , du = 1/2(1/t)dt?

 

[judahs_lion]

actually du = 1/t

 

[judahs_lion]

that would make give me 1/2lnt (1/t), which is udu +C. so it breaks down to u^2/2 + C?

 

[MysticDude]

actually du = 1/t

Actually no. I even checked with Wolfram Alpha. The derivative of $$\frac{1}{2} * ln(t) = \frac{1}{2t}$$

http://www.wolframalpha.com/input/?i=deriv+(1/2)ln(x)"

 

[Dick]

Yea i think i integrated rather then differentiated to get du. so u = 1/2lnt , du = 1/2(1/t)dt?

Sure. You could have also done it with ln(sqrt(t)). (ln(sqrt(t))'=(1/sqrt(t))*(dsqrt(t)/dt)=(1/sqrt(t))*(1/(2*sqrt(t)))=1/(2t). That's the chain rule. BTW if you aren't going to use TeX you should probably use a few more parentheses. 1/2lnt could mean (1/2)ln(t) or 1/(2ln(t)) etc.

 

[MysticDude]

Sure. You could have also done it with ln(sqrt(t)). (ln(sqrt(t))'=(1/sqrt(t))*(dsqrt(t)/dt)=(1/sqrt(t))*(1/(2*sqrt(t)))=1/(2t). That's the chain rule. BTW if you aren't going to use TeX you should probably use a few more parentheses. 1/2lnt could mean (1/2)ln(t) or 1/(2ln(t)) etc.

Okay I'll think about that next time.

 

[judahs_lion]

Actually no. I even checked with Wolfram Alpha. The derivative of $$\frac{1}{2} * ln(t) = \frac{1}{2t}$$

http://www.wolframalpha.com/input/?i=deriv+(1/2)ln(x)"

Is that implicit differentiation?

 

[MysticDude]

Is that implicit differentiation?

No all it is is just normal differentiation. See all you have to do if factor out the constant( 1/2 in this case). So it would look like $$(\frac{1}{2}) * (\frac{d}{dt})(ln(t)) = \frac{1}{2t}$$

 

[judahs_lion]

No all it is is just normal differentiation. See all you have to do if factor out the constant( 1/2 in this case). So it would look like $$(\frac{1}{2}) * (\frac{d}{dt})(ln(t)) = \frac{1}{2t}$$

but i thought the 1/2 would be eliminated because it is a constant

 

[MysticDude]

but i thought the 1/2 would be eliminated because it is a constant

If you can factor out a constant, then do so. Plus if you look at the link that I gave you. There should be a "Show Steps" link. It says to factor out the constant.

AND even if you do it by using the product rule (I'm doing this so you can see that the 1/2 stays)

$$(\frac{1}{2}[(\frac{d}{dt})(ln(t))] + (\frac{d}{dt})(\frac{1}{2})(ln(t))$$
We have $$\frac{1}{2}*\frac{1}{t} + 0(this-is-the-derivative-of \frac{1}{2}) * ln(t)$$
so this shows that the derivative of $$\frac{1}{2} * ln(t) = \frac{1}{2t}$$I hope you understand!

 

[judahs_lion]

If you can factor out a constant, then do so. Plus if you look at the link that I gave you. There should be a "Show Steps" link. It says to factor out the constant.

AND even if you do it by using the product rule (I'm doing this so you can see that the 1/2 stays)

$$(\frac{1}{2}[(\frac{d}{dt})(ln(t))] + (\frac{d}{dt})(\frac{1}{2})(ln(t))$$
We have $$\frac{1}{2}*\frac{1}{t} + 0(this-is-the-derivative-of \frac{1}{2}) * ln(t)$$
so this shows that the derivative of $$\frac{1}{2} * ln(t) = \frac{1}{2t}$$


I hope you understand!

Thank you

 

[Dick]

Thank you

Ok, so after all that fuss, you can do the integral, right?

 

[judahs_lion]

Ok, so after all that fuss, you can do the integral, right?

not really. I am thinking insted of it substituting to udu, it would be something like u((1/2)du)?

 

[Dick]

not really. I am thinking insted of it substituting to udu, it would be something like u((1/2)du)?

Argh. Why don't you use the original substitution you suggested, u=ln(t^(1/2)). It works great. I promise you.

 

[╔(σ_σ)╝]

not really. I am thinking insted of it substituting to udu, it would be something like u((1/2)du)?

Argh. Why don't you use the original substitution you suggested, u=ln(t^(1/2)). It works great. I promise you.

I promise too :).

 

[judahs_lion]

I just don't see it. Don't see du inside the original function.

 

[Dick]

I just don't see it. Don't see du inside the original function.

I'm going to have to ask you again. If u=ln(t^(1/2)), what's du? Why don't you see du (or some constant multiple of it) in the original function?

 

[judahs_lion]

I'm going to have to ask you again. If u=ln(t^(1/2)), what's du? Why don't you see du (or some constant multiple of it) in the original function?

du = 1/(2t). Only thing i can think of is bringing the 1/2 down but don't see how that helps me.

 

[Dick]

du = 1/(2t). Only thing i can think of is bringing the 1/2 down but don't see how that helps me.

It's du=(1/(2t))*dt, ok? Can you show what you get when you use that substitution into the original integral?

 

[judahs_lion]

2(udu) ?

 

[MysticDude]

2(udu) ?

I think that that's supposed to be $$2\int udu$$

 

[judahs_lion]

I think that that's supposed to be $$2\int udu$$

yea, i don't know have to use symbols in here.

 

[Dick]

yea, i don't know have to use symbols in here.

Ok, so you can integrate that, right?

 

[MysticDude]

I don't want to tell you the answer, but I'll give you a clue...that will pretty much help you get the answer. Now that we have everything with "u", we can integrate normally. As if you were integrating plain x. All you have to do is use the reverse power rule.

So what would you get if you use the reverse power rule on u?

 

[fizzynoob]

its a very simple integral after you use U substitution


$$\frac{1}{2}$$$$\int$$$$\frac{ln(t)}{t}$$dt

u = ln(t)
du = ?

 

[judahs_lion]

Ok, so you can integrate that, right?

2[{u^(3/2)}/(3/2)}]

so

[lnt^2] /3 ?

 

[MysticDude]

2[{u^(3/2)}/(3/2)}]

so

[lnt^2] /3 ?

from where did you get the 3/2. Remember that ln(√t) does not equal (ln(t))^(1/2). I think that the answer should have been u² => (ln(√t))².


I'm not 100% on this, but I think that this is the correct one.