Solve Integrals & Prove Continuousness: Pi/2 Solution

[BartTheMan]

Homework Statement

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

Homework Equations

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The Attempt at a Solution

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

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

 

[Mark44]

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

Homework Equations

/

The Attempt at a Solution

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

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

I need to run now, but will take a closer look at this later this afternoon.

 

[Dick]

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.

 

[BartTheMan]

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.

 

[BartTheMan]

Problem solved, found the answer

 

[Dick]

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.