• Support PF! Buy your school textbooks, materials and every day products Here!

Integration using Trig Substitution

  • Thread starter Brickster
  • Start date
  • #1
21
0

Homework Statement



[tex]\int \frac{cosx dx}{\sqrt{1 + sin^{2}x}}[/tex]

Homework Equations



Expression: [tex]\sqrt{a^{2} + x^{2}}[/tex]
Substitution: x = a*tan[tex]\Theta[/tex]
Identity: 1 + tan[tex]^{2}\Theta = sec^{2}\Theta[/tex]

The Attempt at a Solution


I have tried using Trig Substitution, but I end up getting an equation much like the one I started only it contains secants and tangents instead of cosine and sine. For some reason I only see a circular route that would just take me back to the original equation.
 
Last edited:

Answers and Replies

  • #2
cristo
Staff Emeritus
Science Advisor
8,107
73
Try the substitution y=sin(x). That should then put the integrand in a familiar form.
 
  • #3
21
0
as in changing the bottom from 1 + (sinx)^2 to 1 + y^2 ?


I can get to:

[tex]\int \frac{sec^{2}\Theta d\Theta}{\sqrt{1+tan^{2}\Theta}}[/tex]

But after that it makes no sense, I just get back to the original equation when I sub. again.
 
Last edited:
  • #4
Gib Z
Homework Helper
3,346
5
After y=sin x, and then a tan theta = y, you should have the same integrand with a plus sign, not a minus. Then use the Pythagorean identities to simplify the square root.
 
  • #5
21
0
So I should end up with:

[tex]\int sec\Theta d\Theta}[/tex]

Correct?

then I would get ln [tex]\left| sec\Theta+tan\Theta \right|[/tex] + C
 
Last edited:
  • #6
cristo
Staff Emeritus
Science Advisor
8,107
73
If you use the substitution I suggested, then you end up with [tex]\int\frac{dy}{\sqrt{1+y^2}}[/tex]. This is a standard integral, is it not?
 
  • #7
21
0
If you use the substitution I suggested, then you end up with [tex]\int\frac{dy}{\sqrt{1+y^2}}[/tex]. This is a standard integral, is it not?
Yes, I'm fairly certain that is a standard integral, and it comes out to:

[tex]sin^{-1}y + C[/tex]

But I'm also fairly certain that I am supposed to solve this problem using the expression, substitution, and identity when I do the trig sub. Maybe I am overcomplicated the problem and should just stick to the arcsin answer, Gib Z's method seems to be the one I'm after
 
Last edited:
  • #8
cristo
Staff Emeritus
Science Advisor
8,107
73
Yes, I'm fairly certain that is a standard integral, and it comes out to:

[tex]sin^{-1}\frac{y} + C[/tex]
Close; the solution is [tex]\sinh^{-1} y +C[/tex]
 
  • #9
21
0
Close; the solution is [tex]\sinh^{-1} y +C[/tex]
My mistake, you're right. Thanks!
 
Last edited:
  • #10
Gib Z
Homework Helper
3,346
5
It's always nice to know alternative methods, because we can know express the inverse hyperbolic sine function in terms of the natural logarithm =] Which we could have also done more directly, but o well!
 

Related Threads on Integration using Trig Substitution

Replies
3
Views
603
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
835
  • Last Post
Replies
6
Views
3K
Replies
4
Views
937
Replies
5
Views
6K
Replies
8
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
13
Views
1K
Top