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.