Polar Arc Length Question

1. Jun 23, 2010

stau40

1. The problem statement, all variables and given/known data
Find the length of the cardioid with equation r = 1 + cos (theta) located in the first quadrant

2. Relevant equations
f (theta) = 1 + cos (theta) f'(theta) = -sin (theta) s = antiderivative (0 to (pi/2)) sq rt (f(theta)^(2) + f'(theta)^(2)) d(theta)

3. The attempt at a solution
I have worked thru the problem and have arrived at sq rt (2) antiderivative (0 to (pi/2)) sq rt (1-sin (theta)) d(theta) but have hit a road block here. Is there a trig identity that will help solve this? I haven't had any luck finding anything and don't know how else to proceed. Thanks in advance!

2. Jun 23, 2010

rock.freak667

$$1-sin x \times \frac{1+sinx}{1+sinx} = ?$$

then use cos2x+sin2x = 1

then a trig substitution.

3. Jun 23, 2010

stau40

It's been a long day, but I don't understand where the (1+sin x)/(1+sin x) comes from? From what I see that equals 1 and again leaves me with 1-sin x?

4. Jun 23, 2010

Staff: Mentor

Yes, 1 + sinx over itself equals 1, and you can always multiply by 1. If you carry out the multiplication suggested by rock.freak667, what do you get?

5. Jun 23, 2010

rock.freak667

multiply it by 1-sinx and then what would you get? (you are multiplying by '1' so that you are not changing the integral)

6. Jun 23, 2010

stau40

I'm getting ((1-sinx)*(1+sinx))/1+sinx = (1+sinx-sinx-sinx^(2))/(1+sinx) = (1-sinx^(2))/(1+sinx) = (cosx^(2))/(1+sinx). Is this correct, or did I mess it up?

7. Jun 23, 2010

rock.freak667

Yes that is correct. So now instead of integrating √(1-sinx) wrt x, you can now integrate

$$\int \sqrt{\frac{cos^2x}{1+sinx}} dx$$

now remember that √(a/b) = √a/√b

So what does the new integral become?

8. Jun 23, 2010

stau40

I'm getting the antiderivative of (cosx)/(1+(sq rt sinx))

9. Jun 23, 2010

Bohrok

$$\sqrt{1 + \sin x} \neq 1 + \sqrt{\sin x}$$

10. Jun 23, 2010

stau40

$$\int {\frac{cos~x}{\sqrt{1+sin~x}}} dx$$