- #1

- 1,462

- 44

My question is, what if I just didn't happen to catch that symmetry argument? Would I just not be able to use the half-angle substitution, since both bounds go to zero? Is noticing the symmetry really the only way around this?

- I
- Thread starter Mr Davis 97
- Start date

- #1

- 1,462

- 44

My question is, what if I just didn't happen to catch that symmetry argument? Would I just not be able to use the half-angle substitution, since both bounds go to zero? Is noticing the symmetry really the only way around this?

- #2

haushofer

Science Advisor

- 2,479

- 865

- #3

haushofer

Science Advisor

- 2,479

- 865

So, for which t one has arctan(t)=pi?

I'd say it lies outside of arctan's range.

I'd say it lies outside of arctan's range.

- #4

- 294

- 146

I tried this substitution and found it intractable. For the integral,I want to make the tangent half-angle substitution t=tan(x/2)

$$I=2\int_ {0}^{\pi}\frac {1-cos(x)} {3+ cos(x)}dx$$

Let ## t=sin(\frac {x} {2})## with ##dt = \frac {1}{2}t \sqrt {1-t^2}dx## and the integral becomes,

$$I= 2\int_{0}^{1} \frac {tdt}{\sqrt {1-t^2}(1-\frac {t^2} {2})}$$

Make another substitution with ##u=t^2## and ##du=2tdt## to get,

$$I=\int_{0}^{1}\frac {du}{\sqrt {u}(1- \frac {u} {2}) \sqrt {1-u}}$$

I compare this with the hypergeometric integral

$$B(b,c-b)\;_ {2}F_1(a,b;c;z)=\int_{0}^{1}x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a}dx$$

where B is the beta function. I get ##a=1##, ##b= \frac {1}{2}##, ##c=1##, ##z=\frac {1}{2}##, ##B(\frac {1}{2},\frac {1}{2}) = \pi##, to find

$$I=\pi \;_ {2}F_1(1,\frac {1}{2};1;\frac {1}{2})$$

I suspect there is a formula for giving this hypergeometric function as an algebraic number but I don't have access to the literature quoted on the Wikipedia page for the hypergeometric function.

Peace,

Fred

- #5

- 1,462

- 44

Good point. I guess my question now is, if I have an integral ##\int_a^b f(x) ~dx##, and I want to make the change of variables ##u = g(x)## or ##x = h(u)##, what are the formal restrictions on ##h## and ##g##?So, for which t one has arctan(t)=pi?

I'd say it lies outside of arctan's range.

- #6

- 11,356

- 3,969

The domain of ##g(x)## (or equivalently, the codomain of ##h(u)##) must contain the interval ##(a,b)##.Good point. I guess my question now is, if I have an integral ##\int_a^b f(x) ~dx##, and I want to make the change of variables ##u = g(x)## or ##x = h(u)##, what are the formal restrictions on ##h## and ##g##?

- Replies
- 3

- Views
- 674

- Replies
- 2

- Views
- 2K

- Replies
- 4

- Views
- 6K

- Replies
- 7

- Views
- 17K

- Replies
- 2

- Views
- 3K

- Replies
- 11

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 8K

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 2K