Another Integral Problem

  • Thread starter Temp0
  • Start date
  • #1
79
0

Homework Statement



http://i.imgur.com/u1De0i3.png

Homework Equations





The Attempt at a Solution



So I notice that the bottom is in the form x^2 - a^2 where a = 7, so I use trig substitution to start this off.

x = 7secθ, dx = 7secθtanθ, and finally, x^2 - 49 = 49sec(θ)^2 - 49 = 49 tan^2(θ)
Substituting into the integral, I get
7∫(sec^2(θ) - secθ) dθ, which basically turns into:
7tanθ - 7 ln |secθ + tanθ|.
After putting x back into the equation, I end up with:
√(x^2-49) - 7 ln|(x/7) + (√(x^2-49)/7)| + C
I would just like your help in checking my answers, because I don't get any of the answers provided in the multiple choice, and i'm always hesitant to pick "none of the above". Thank you.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,263
619

Homework Statement



http://i.imgur.com/u1De0i3.png

Homework Equations





The Attempt at a Solution



So I notice that the bottom is in the form x^2 - a^2 where a = 7, so I use trig substitution to start this off.

x = 7secθ, dx = 7secθtanθ, and finally, x^2 - 49 = 49sec(θ)^2 - 49 = 49 tan^2(θ)
Substituting into the integral, I get
7∫(sec^2(θ) - secθ) dθ, which basically turns into:
7tanθ - 7 ln |secθ + tanθ|.
After putting x back into the equation, I end up with:
√(x^2-49) - 7 ln|(x/7) + (√(x^2-49)/7)| + C
I would just like your help in checking my answers, because I don't get any of the answers provided in the multiple choice, and i'm always hesitant to pick "none of the above". Thank you.

One of those answers differs from your answer by a constant.
 
  • #3
79
0
What do you mean? Hmm, I can't really see any way to rearrange it like that, I think I know the one you're talking about though.
 
  • #4
Dick
Science Advisor
Homework Helper
26,263
619
What do you mean? Hmm, I can't really see any way to rearrange it like that, I think I know the one you're talking about though.

For example, log(x/7)=log(x)-log(7).
 
  • #5
79
0
Ohhhh! Let's see if I can go any further now, thanks alot.
 

Related Threads on Another Integral Problem

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
2
Views
741
  • Last Post
Replies
7
Views
1K
Replies
1
Views
1K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
3
Views
972
  • Last Post
Replies
15
Views
1K
  • Last Post
Replies
1
Views
831
Top