# 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

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