1. ## simple intergration help

ok
what is the integral of ( 1/2t)

i did a u sub

so u = 2t
du = 2

2 * ln2t

is this correct?

2. $\displaystyle \int \frac{1}{2t} \ dt \ = \ \frac{1}{2} \int \frac{1}{t} \ dt$

which is a standard integral.

3. Originally Posted by o_O
$\displaystyle \int \frac{1}{2t} \ dt \ = \ \frac{1}{2} \int \frac{1}{t} \ dt$

which is a standard integral.
ah so it is lnt/2

?

4. yeah it is.. ty sir.

5. No. You will get 1/2 ln(t).
They are not the same.
1/2 ln(t) <> ln(t/2). They maybe different.

6. Originally Posted by oswaldo
No. You will get 1/2 ln(t).
They are not the same.
1/2 ln(t) <> ln(t/2). They maybe different.
ok but what am i doing wrong then

int (1/2t)

so if u just do a u sub instead of pulling the 1/2 out.

u = 2t
du = 2du

(1/2) int [(2du)/(2t)

=

1/2 lnu

1/2 * ln 2t

why is it 1/2 * lnt and not 1/2 * ln2t

what did i do wrong?

7. You don't need any kind of u or whatever conversion. It is a simple basic integral. All you need to know is:
int(1/t) = ln(t). THAT'S IT.

1/2 is just a constant. So the answer of your questions is:
int(1/2t)=(1/2) int(1/t)=(1/2) ln(t). AND THAT'S IT.
Which may be different than ln(t/2).

a*ln(t)=ln(a*t) is NOT always true.

So (1/2)*ln(t) = ln((1/2)*t) is NOT always true.

-O

8. If for some reason you decided to use a u-substitition:

$\displaystyle {\color{red}u = 2t} \ \Rightarrow \ {\color{blue}\tfrac{1}{2}du = dt}$

We then get: $\displaystyle \int \frac{1}{{\color{red}2t}} \ {\color{blue}dt} \ = \ \frac{1}{2} \int \frac{{\color{blue}du}}{{\color{red}u}} \ = \ \frac{1}{2} \ln u + C \ = \ \frac{1}{2} \ln 2t + C$

Looks a bit different. But recall: $\displaystyle \ln (ab) = \ln a + \ln b$

So: $\displaystyle \frac{1}{2} \ln 2t + C = \frac{1}{2} \ln 2 + \frac{1}{2} \ln t + C$

But $\displaystyle \tfrac{1}{2}\ln 2$ is nothing more than a constant. So we'll combine it with $\displaystyle C$ to make a new constant of integration which we'll call $\displaystyle K$.

i.e. $\displaystyle \int \frac{1}{2t} \ dt \ = \ \frac{1}{2} \ln 2t + C \ = \ \frac{1}{2} \ln t + K$

In other words, $\displaystyle \tfrac{1}{2} \ln 2t$ and $\displaystyle \tfrac{1}{2} \ln t$ are two legit antiderivatives of $\displaystyle \tfrac{1}{2t}$