# Find the length of a curve

#### renyikouniao

##### Member
find the length of a curve given by:
f(x)=integral(upper bound: x lower bound: 0) (cos^2(x)+4cos(x)+1)^0.5 dx

Here's my solution:
I use the equation L=integral ( upper bound: a lower bound: b ) sqrt[1+(f'(x))^2]

f'(x)=(cos^2(x)+4cos(x)+1)^0.5

1+[f'(x)]^2=2+cos^(x)+4cos(x)=[cos(x)+2]^2-2

L=integral( upper bound: x lower bound: 0 ) [2+cos^2(x)+4cos(x)]^0.5

This is where I am stuck...Am I right so far?

#### MarkFL

Staff member
It is probably best to let the dummy variable of integration be something other than $x$...most times I see $t$ being used in such cases:

$$\displaystyle f(x)=\int_0^x \sqrt{\cos^2(t)+4\cos(t)+1}\,dt$$

However, I am unclear what the limits of integration are to be when computing the arc-length along this curve. If you are to use parameters, I suggest something other than $x$, such as $a$ and $b$, and then we could give the arc-length $s$ as:

$$\displaystyle s=\int_a^b\sqrt{1+\left(f'(x) \right)^2}\,dx$$

You have correctly applied the derivative form of the FTOC to obtain:

$$\displaystyle f'(x)=\sqrt{\cos^2(x)+4\cos(x)+1}$$

and thus:

$$\displaystyle 1+\left(f'(x) \right)^2=1+\cos^2(x)+4\cos(x)+1=\cos^2(x)+4\cos(x)+2$$

and so we have:

$$\displaystyle s=\int_a^b\sqrt{\cos^2(x)+4\cos(x)+2}\,dx$$

Completing the square as you did, gives:

$$\displaystyle s=\int_a^b\sqrt{(\cos(x)+2)^2-2}\,dx$$

At this point, I might consider the substitution:

$$\displaystyle \cos(x)+2=\sqrt{2}\tan(\theta)$$

but, this leads to an integrand that seem far more complicated.

Are you certain you have given all the relevant information?

#### renyikouniao

##### Member
It is probably best to let the dummy variable of integration be something other than $x$...most times I see $t$ being used in such cases:

$$\displaystyle f(x)=\int_0^x \sqrt{\cos^2(t)+4\cos(t)+1}\,dt$$

However, I am unclear what the limits of integration are to be when computing the arc-length along this curve. If you are to use parameters, I suggest something other than $x$, such as $a$ and $b$, and then we could give the arc-length $s$ as:

$$\displaystyle s=\int_a^b\sqrt{1+\left(f'(x) \right)^2}\,dx$$

You have correctly applied the derivative form of the FTOC to obtain:

$$\displaystyle f'(x)=\sqrt{\cos^2(x)+4\cos(x)+1}$$

and thus:

$$\displaystyle 1+\left(f'(x) \right)^2=1+\cos^2(x)+4\cos(x)+1=\cos^2(x)+4\cos(x)+2$$

and so we have:

$$\displaystyle s=\int_a^b\sqrt{\cos^2(x)+4\cos(x)+2}\,dx$$

Completing the square as you did, gives:

$$\displaystyle s=\int_a^b\sqrt{(\cos(x)+2)^2-2}\,dx$$

At this point, I might consider the substitution:

$$\displaystyle \cos(x)+2=\sqrt{2}\tan(\theta)$$

but, this leads to an integrand that seem far more complicated.

Are you certain you have given all the relevant information?
Yes,and this is where I am stuck..