Hi everyone,
Can someone give me a hint on this one, please?
I have the integral from 1 to the square root of 3 arctan(1/x) dx
I said that u=arctan(1/x)
du=????
dv=???????
v=???????
I'm not sure if my u is correct or not. I don't think it is because now I can't choose anything for du, right? Should it be u=arctan and du=1/x????
Thank you
This is what I got:
u=arctan(1/x)
du=1/1+x^2dx
dv=dx
v=x
integral 1 to the square root of 3 arctan(1/x)dx=xtan^-1(1-x)]one to the square root of 3-integral 1 to the square root of 3 x/1+x^2dx=
square root of 3tan^-1(1-sqauare root of 3)-[1(tan^-1)(1-1)]-integral 1 to the square root of three x/1+x^2dx=
-1.095-0-integral 1 to the square root of three x/1+x^2dx
-1.095-the integral 0 to the square root of 9x/1+x^2dx
t=1+x^2
dt=2xdx
xdx=1/2dt
when x=square root of 3 t=4
when x=1 t=2
integral 1 to square root of 3 x/1+x^2dx=1/2 integral 2 to 4 dt/t=1/2ln(t)]2 to 4=
1/2(ln4-ln2)=1/2ln2=
integral 1 to square root of 3 tan^-1xdx=-1.095-integral 1 to square root 3 x/1+x^2=
-1.095-1/2ln2
Can you tell me if this is correct and if it isn't where I made my mistake?
Thank you very much
by the way, chocolatelover, writing out sentences to describe mathematical symbols and such makes your post tedious to read, at least for me. if you can't use LaTex, try these shorthands, it will make your posts more readable AND save you a lot of typing
instead of: the square root of x
type: sqrt(x)
instead of: the integral of xdx
type: int(x)dx
instead of: the integral from a to b of xdx
type: int_{a}^{b} (x)dx
OR
int (x)dx ...from a to b (keep the limits at the end of the line after you type everything else)
example:
instead of typing, as you did: integral 1 to the square root of 3 arctan(1/x)dx=xtan^-1(1-x)]one to the square root of 3-integral 1 to the square root of 3 x/1+x^2dx
you could type: int_{1}^{sqrt(3)}arctan(1/x)dx = xarctan(1/x) - int_{1}^{sqrt(3)}[x/(1+x^2)]dx
it is understandable that the limits for xarctan(1/x) is from 1 to sqrt(3), if you want to emphasis it, leave off the limits to the end
int [arctan(1/x)]dx = xarctan(1/x) - int [x/(1+x^2)]dx ...from 1 to sqrt(3)
now, see how nicer and shorter that is?
Or better yet, you could learn to use LaTex, it's not as hard as you may think. See our LaTex Tutorial
Hi,
"arctan" is the same thing as tan^-1 right?
Can you please explain what you did? How did you get + integral? Isn't the formula [tex]\int[\math] a to b g'(x)dx=f(x)g(x)]a to b - [tex]\int[\math] a to be g(x)f'(x)dx
How did you get the x for =xarctan? Isn't that supposed to be f(x) wich is arctan(1/x)?
Where you said "we use the substitution u: 1+x^2" is du=x
dv=2x
v=x^2/2
To finish the problem do I need to plug in 1 and [tex]sqrt{x}[\math] to every x and subtract the two?
Thank you
again, you put the wrong slash on the [tex] tag! please follow instructions carefully when we type them, we don't do it just for fun, we do it to help you
yes
the formula has a minus sign in it, yes, but du is negative, a negative times a negative is positiveCan you please explain what you did? How did you get + integral? Isn't the formula [tex]\int[\math] a to b g'(x)dx=f(x)g(x)]a to b - [tex]\int[\math] a to be g(x)f'(x)dx
The formula for integration by parts is:How did you get the x for =xarctan? Isn't that supposed to be f(x) wich is arctan(1/x)?
and
thus
no. integration by substitution is different from integration by parts, there is no need to find dv and v when doing substitution
Where you said "we use the substitution u: 1+x^2" is du=x
dv=2x
v=x^2/2
you plug in first and then minus the answer you get when you plug in 1To finish the problem do I need to plug in 1 and [tex]sqrt{x}[\math] to every x and subtract the two?
Thank you
this is what the Fundamental Theorem of Calculus says. It's a very important theorem, please look it up in your text, you must know it