YAMT: If anyone wants do to integrals at night

Ricemarine

Lifer
Sep 10, 2004
10,507
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I am at a loss as to how to go about this equation

e^t * sqrt[25 - e^(2t)] dt

if u = e^t
du = e^t dt
dv = sqrt[25 - e^(2t)dt
v = 25/2 sin^-1 (e^x/5)

but if I plug v into the equation uv = int[v du], it's really messed up...
Wanna give a hand night shift? :)
 

LongCoolMother

Diamond Member
Sep 4, 2001
5,675
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Are you sure you need to do integration by parts?

How about u=e^t
du = e^t dt

so it becomes

sqrt(25-u^2) du

and then trig substitution?

I could be wrong, im like half asleep and I didnt write it on paper or anything. just going off the top of my head.
 

chuckywang

Lifer
Jan 12, 2004
20,133
1
0
Second of all, you're on the wrong track to begin with. This problem requires a bunch of steps.

1) First let u = e^t, so du=e^t dt. Substituting, your integral now becomes sqrt(25-u^2) du.

2) Rewrite this integral as (25-u^2)/(sqrt(25-u^2)) du = 25/sqrt(25-u^2) du - u^2/(sqrt(25-u^2)) du

The integral of the first term is an Arcsin. The integral of the second term can be found either by integration by parts or trig substitution (let u = 5sin(theta)).
 

Ricemarine

Lifer
Sep 10, 2004
10,507
0
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Originally posted by: chuckywang
Second of all, you're on the wrong track to begin with. This problem requires a bunch of steps.

1) First let u = e^t, so du=e^t dt. Substituting, your integral now becomes sqrt(25-u^2) du.

2) Rewrite this integral as (25-u^2)/(sqrt(25-u^2)) du = 25/sqrt(25-u^2) du - u^2/(sqrt(25-u^2)) du

The integral of the first term is an Arcsin. The integral of the second term can be found either by integration by parts or trig substitution (let u = 5sin(theta)).

Ohhhhhhhhhhhhhh.................
Hum... I knew I went about it the wrong way...

Thanks chuckywang :cookie: :)
 

Dangerer

Golden Member
Mar 15, 2005
1,128
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Originally posted by: compuwiz1
Nice read! This is the geeky stuff AT was born from. :)

what? this is just basic calculus that most students learn in high school o_O