Integration problem (need help)

  • #1
if a spaceship accelerates, from restat time t = 0, at a rate of 2t / sqrt(1 + t^2) m/s at time t, calculate in years to 1 significant figure how long it would take to reach the speed of light?

my attempt

speed of light = 3.0 x 10^8 m/s
integrate the rate function 2t / sqrt(1 + t^2) from t = 0, to the t we are looking for

integrating 2t / sqrt(1 + t^2)

using integration by parts;

let u = 1 + t^2, then du = 2t dt

WHERE DO I GO FROM HERE
 

Answers and Replies

  • #2
35,226
7,046
Before you go at an integral using integration by parts, you should always see if a simpler substitution will work. What you have shown is exactly the substitution I would use (your work does not show that you are doing integration by parts).

Using this substitution, what does your new integral look like?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
966
if a spaceship accelerates, from restat time t = 0, at a rate of 2t / sqrt(1 + t^2) m/s at time t, calculate in years to 1 significant figure how long it would take to reach the speed of light?

my attempt

speed of light = 3.0 x 10^8 m/s
integrate the rate function 2t / sqrt(1 + t^2) from t = 0, to the t we are looking for

integrating 2t / sqrt(1 + t^2)

using integration by parts;

let u = 1 + t^2, then du = 2t dt

WHERE DO I GO FROM HERE

That is NOT "integration by parts". Perhaps you should review that.
You want to integrate
[tex]\int \frac{2t dt}{\sqrt{1+ t^2}}[/tex]
and you say [itex]u= 1+ t^2[/itex] and [itex]du= 2tdt[/itex]. Okay, doesn't it make sense to replace the "2tdt" in the integral by du and the "[itex]1+ t^2[/itex] in the integral by u?
You might want to remember that [itex]1/\sqrt{a}= a^{-1/2}[/itex].
 
  • #4
cheers, my mistake
 

Related Threads on Integration problem (need help)

Replies
1
Views
815
  • Last Post
Replies
7
Views
1K
H
Replies
14
Views
9K
  • Last Post
Replies
6
Views
1K
Replies
1
Views
539
  • Last Post
Replies
2
Views
997
  • Last Post
Replies
1
Views
906
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
4
Views
1K
Top