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Proof integrals

  • Thread starter BartTheMan
  • Start date
  • #1

Homework Statement


a) Solve:[tex]^{Pi}_{0}[/tex][tex]\int[/tex][tex]\frac{sin(x)}{1 + cos²x}[/tex]dx
b) Proof that for each f, continuous in [0, a], [tex]^{a}_{0}[/tex][tex]\int[/tex][tex]{f(x)}[/tex]dx = [tex]^{a}_{0}[/tex][tex]\int[/tex][tex]{f(a-x)}[/tex]dx
c) Use a and b to solve [tex]^{Pi}_{0}[/tex][tex]\int[/tex][tex]\frac{x sin(x)}{1 + cos²x}[/tex]dx

Homework Equations


/

The Attempt at a Solution


a) t = cos(x)
dt/dx = sin(x)
dt = sin(x)*dx
[tex]^{Pi}_{0}[/tex][tex]\int[/tex][tex]\frac{sin(x)}{1 + cos²x}[/tex]dx
= [tex]^{1}_{-1}[/tex][tex]\int[/tex][tex]\frac{dt}{1 + t²}[/tex]dt
= arctan(1)-arctan(-1) = Pi/2

b) t = a - x
dt/dx = -1
-dt = dx
[tex]^{0}_{a}[/tex][tex]\int[/tex][tex]{-f(t)}[/tex]dt
= [tex]^{a}_{0}[/tex][tex]\int[/tex][tex]{f(t)}[/tex]dt
= [tex]^{a}_{0}[/tex][tex]\int[/tex][tex]{f(x)}[/tex]dx

c) I have no idea to start this should I replace x with Pi-x, I tried this but i'm not getting any further
 

Answers and Replies

  • #2
33,162
4,846
Cleaned up your LaTeX.
Tips:
1. Use only one pair of [ tex] and [ /tex] tags per line. Your stuff will format much more nicely that way.
2. For a definite integral, do it like this:
[ tex]\int _{0} ^ {\pi} f(x) dx [ /tex]
3. Don't use small font superscripts inside LaTeX code. They don't get rendered. For example, I didn't realize that you had 1 + cos2x in the denominator until I looked at your LaTeX.
4. For Greek letters, precede them with a backslash; e.g., \pi, \alpha, etc. Note that the names are case-sensitive. If you capitalize the name (like \Pi), you get the upper-case form of the letter.

Homework Statement


a) Solve:[tex]\int ^{\pi}_{0} \frac{sin(x)dx}{1 + cos^2(x)}[/tex]
b) [STRIKE]Proof[/STRIKE] Prove that for each f, continuous in [0, a], [tex]\int ^{a}_{0}f(x)dx = \int ^{a}_{0}f(a-x) dx[/tex]
c) Use a and b to solve [tex]\int ^{\pi}_{0}\frac{x sin(x)}{1 + cos^2(x)}dx[/tex]

Homework Equations


/

The Attempt at a Solution


a) t = cos(x)
dt/dx = sin(x)
dt = sin(x)*dx
[tex]\int^{\pi}_{0}\frac{sin(x)}{1 + cos^2x}dx[/tex]
= [tex]\int^{1}_{-1}\frac{dt}{1 + t^2}dt[/tex]
= arctan(1)-arctan(-1) = Pi/2

b) t = a - x
dt/dx = -1
-dt = dx
[tex]\int^{0}_{a}-f(t)dt[/tex]
= [tex]\int ^{a}_{0}f(t)dt[/tex]
= [tex]\int^{a}_{0}f(x) dx[/tex]

c) I have no idea to start this should I replace x with Pi-x, I tried this but i'm not getting any further
I need to run now, but will take a closer look at this later this afternoon.
 
Last edited:
  • #3
Dick
Science Advisor
Homework Helper
26,258
618
Yes, replace x by pi-x and equate the two integrals. Then solve for the integral you are after. Thanks for unveiling the cos(x)^2, Mark44, without the square both integrals diverge.
 
  • #4
Yes, replace x by pi-x and equate the two integrals. Then solve for the integral you are after. Thanks for unveiling the cos(x)^2, Mark44, without the square both integrals diverge.
Sorry for my bad Latex use

Sorry, but I don't really understand which integrals I have to equate.
 

Attachments

  • #5
Problem solved, found the answer
 
  • #6
Dick
Science Advisor
Homework Helper
26,258
618
Sorry for my bad Latex use

Sorry, but I don't really understand which integrals I have to equate.
You have two integrals of x*sin(x)/(1+cos(x)^2). Combine them. You know how to do the pi*sin(x)/(1+cos(x)^2) part.
 

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