Integration by parts and u-sub problem

[aria]

I'm having a hard time getting this equation for some reason.
The section we're currently covering is integration by parts, but the teacher suggests that we also use u-substitution for this problem.

int_0^1 (r^3)/sqrt(4+r^2) dr

the teacher suggested: int_0^1 (r^2)*r/sqrt(4+r^2) dr
u = r^2
dv = r/sqrt(4+r^2)
and that we find v by u substitution.

However, I kind of get confused on how that's supposed to work with integration by parts, and get everything all mixed up.

Using m, instead of u for a variable since that one is already in use I get...

dv=r/sqrt(4+r^2) = r*1/sqrt(4+r^2)
m= 4+r^2

dm/dx= 2r
1/2 dm = r dx

integrating m => m^(-1/2)
2m^(1/2)

changing dv= r*m^(-1/2) = (1/2)m^(-1/2)

so, I think that this would give me:

int_0^1 (r^3)/sqrt(4+r^2) dr = (r^2)*2m^(1/2)| - (1/2) int_0^1 2m^(1/2) dm

={ [(1^2)*2(4+1^2)^(1/2)]-0} - (2/3)m^(3/2)
={ [(1^2)*2(4+1^2)^(1/2)]-0} - {[(2/3)*(4+1^2)^(3/2)]-[(2/3)*(4+0)^(3/2)]}
=sqrt(8+2)-[(2/3)*(3/sqrt(5))]+[(2/3)*(3/sqrt(4))]
​

but I'm pretty sure that this is completely wrong

 

[stapel]

I'm having a hard time getting this equation for some reason.
The section we're currently covering is integration by parts, but the teacher suggests that we also use u-substitution for this problem.

int_0^1 (r^3)/sqrt(4+r^2) dr

the teacher suggested: int_0^1 (r^2)*r/sqrt(4+r^2) dr
u = r^2
dv = r/sqrt(4+r^2)
and that we find v by u substitution.

Well, maybe your instructor means that you're to use u-sub in the process of integration by parts. Because using "u" and "dv" (which implies a "du" and a "v") means that you are in fact using integration by parts! (Maybe you should look at inverse trig derivatives...?)

Using m, instead of u for a variable since that one is already in use I get...

dv=r/sqrt(4+r^2) = r*1/sqrt(4+r^2)
m= 4+r^2

dm/dx= 2r
1/2 dm = r dx

Where is "x" coming from?

 

[aria]

Well, maybe your instructor means that you're to use u-sub in the process of integration by parts. Because using "u" and "dv" (which implies a "du" and a "v") means that you are in fact using integration by parts! (Maybe you should look at inverse trig derivatives...?)

I said that the teacher told us to ALSO use u-substitution for the problem, not instead.
The teacher told us that to find v from our dv we had to use u substitution.

I know of no inverse trig derivatives that have a 4 in the denominator.
arcsin' (x) = 1/sqrt(1-x^2) = -arccos' (x) and arctan' (x) = 1/(1+x^2) = - arccot' (x)

Where is "x" coming from?

my bad, I get so used to having an x as a variable that sometimes I just type dx, but that was an obvious typo as x only shows up only in that one spot and no where else, I clearly intended dx to be dr.

 

[Deleted member 4993]

I'm having a hard time getting this equation for some reason.
The section we're currently covering is integration by parts, but the teacher suggests that we also use u-substitution for this problem.

int_0^1 (r^3)/sqrt(4+r^2) dr

the teacher suggested: int_0^1 (r^2)*r/sqrt(4+r^2) dr
u = r^2
dv = r/sqrt(4+r^2)
and that we find v by u substitution.

However, I kind of get confused on how that's supposed to work with integration by parts, and get everything all mixed up.

Using m, instead of u for a variable since that one is already in use I get...

dv=r/sqrt(4+r^2)dr

v = √(4+r^2) → differentiate it to convince yourself

m= 4+r^2

dm/dx= 2r
1/2 dm = r dx

integrating m => m^(-1/2)
2m^(1/2)

changing dv= r*m^(-1/2) = (1/2)m^(-1/2)

so, I think that this would give me:

int_0^1 (r^3)/sqrt(4+r^2) dr = (r^2)*2m^(1/2)| - (1/2) int_0^1 2m^(1/2) dm

={ [(1^2)*2(4+1^2)^(1/2)]-0} - (2/3)m^(3/2)
={ [(1^2)*2(4+1^2)^(1/2)]-0} - {[(2/3)*(4+1^2)^(3/2)]-[(2/3)*(4+0)^(3/2)]}
=sqrt(8+2)-[(2/3)*(3/sqrt(5))]+[(2/3)*(3/sqrt(4))]
​

but I'm pretty sure that this is completely wrong

.
a different method:

use

r = 2 tan(Θ)

dr = 2 sec²(Θ) dΘ

r³/sqrt(4+r^2)dr

8*tan³(Θ)/[2√{1+tan²(Θ)}] * 2 sec²(Θ) dΘ

Now continue...(don't forget to change the limits)