# Polar Arc Length Question

## Homework Statement

Find the length of the cardioid with equation r = 1 + cos (theta) located in the first quadrant

## Homework 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)

## 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!

## Answers and Replies

rock.freak667
Homework Helper
$$1-sin x \times \frac{1+sinx}{1+sinx} = ?$$

then use cos2x+sin2x = 1

then a trig substitution.

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?

Mark44
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?

rock.freak667
Homework Helper
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?

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

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?

rock.freak667
Homework Helper
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?

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?

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

For your denominator,
$$\sqrt{1 + \sin x} \neq 1 + \sqrt{\sin x}$$

How about this: (cosx)/(square root cosx)

Mark44
Mentor
No, not even close. Following rock.freak667's suggestion you should have
$$\int {\frac{cos~x}{\sqrt{1+sin~x}}} dx$$

The reason for his suggestion was to get to an integrand that is relatively easy to antidifferentiate.