Integration help please

  • Thread starter gona
  • Start date
  • #1
9
0
Integrating [tex]\frac{1}{(1+sinx)}[/tex]

i just started learning Integration last week so not exactly sure how to approch this type of question.
 

Answers and Replies

  • #2
537
1
Do you know how to integrate 1/sinx?
 
  • #3
9
0
no this is the first time i have seen where a trigonometric function is on the denominator

but now that i think about it

sinx = [tex]\sqrt{1-cos^2x}[/tex]
and arcsin was equal to [tex]\frac{1}{((1-x^2)}[/tex]

so i guess i could use the U substituition method
and then the answer would be...arcsin(cosx)??? im not too sure
 
Last edited:
  • #4
Dick
Science Advisor
Homework Helper
26,260
619
No, no arcsin. But you might want to try multiplying numerator and denominator by (1-sin(x)). It may look more familiar.
 
Last edited:
  • #5
daniel_i_l
Gold Member
867
0
You could also try the subsitution u = tan(x/2). It looks a little messy but everything cancels out.
 
  • #6
9
0
well im not sure if im doing the right thing but here goes....

sinx = [tex]\frac{2tan(\frac{x}{2})}{1+tan^{2}(\frac{x}{2})}[/tex]

soo then i symplify the equation so that it is

[tex]\frac{1+tan^{2}(\frac{x}{2})}{1+tan^{2}(\frac{x}{2})+2tan(\frac{x}{2})}[/tex]

the i used u = tan[tex](\frac{x}{2})[/tex]

so i get [tex]\frac{1+u^{2}}{(u+1)^{2}}[/tex]

What should i do from here or is this the right way at all?
 
  • #7
Dick
Science Advisor
Homework Helper
26,260
619
It looks to me like you are taking the long way around. Try multiplying numerator and denominator of your original problem by (1-sin(x)). You get (1-sinx)/cos^2(x). If you split that into two integrals you shouldn't have any problem with either of them.
 
  • #8
daniel_i_l
Gold Member
867
0
Dick solution is quicker in this case but to integrate things like
1/(1+cosx+sinx) the substitution u=tan(x/2) is good.
But notice that it isn't x = tan(u/2) but rather u=tan(x/2). In order to get this into something the the example I gave you can show with so trig identities that if u=tan(x/2) then:
dx = 2du/(1+u^2)
sinx = 2u/(1+u^2)
cosx = (1-u^2)/(1+u^2)
If you substitue all that you the (1+u^2)s candel out and you get some rational function which you can solve with the typical rational function method (breaking it into elementry functions...)
 
  • #9
9
0
wow by the way i got the answer using what dick said its actually pretty easy once u break it up...now im gonna try Daniels question gonna see if i can get those now :) thnx alot for the help by the way
the answe was tanx - secx +c
 

Related Threads on Integration help please

  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
3
Views
869
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
7
Views
903
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
20
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
18
Views
2K
Top