Calculus

Integration of Functions

Integration of Functions Logo

Arc Length

  • Arc Length in Rectangular Coordinates

    Let a curve \(C\) be defined by the equation \(y = f\left( x \right),\) where \(f\) is continuous on an interval \(\left[ {a,b} \right].\) We will assume that the derivative \(f^\prime\left( x \right)\) is also continuous on \(\left[ {a,b} \right].\)

    Arc length of a curve
    Figure 1.

    The length of the curve \(y = f\left( x \right)\) from \(x = a\) to \(x = b\) is given by

    \[L = \int\limits_a^b {\sqrt {1 + {{\left[ {f^\prime\left( x \right)} \right]}^2}} dx} .\]

    If we use Leibniz notation for derivatives, the arc length is expressed by the formula

    \[L = \int\limits_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} .\]

    We can introduce a function that measures the arc length of a curve from a fixed point of the curve. Let \({P_0}\left( {a,f\left( a \right)} \right)\) be the initial point on the curve \(y = f\left( x \right),\) \(a \le x \le b.\) Then the arc length of the curve from \({P_0}\left( {a,f\left( a \right)} \right)\) to a point \(Q\left( {x,f\left( x \right)} \right)\) is given by the integral

    \[S\left( x \right) = \int\limits_a^x {\sqrt {1 + {{\left[ {f’\left( t \right)} \right]}^2}} dt} ,\]

    where \(t\) is an internal variable of the integral.

    Arc length function
    Figure 2.

    The function \(S\left( x \right)\) is called the arc length function.

    Arc Length of a Parametric Curve

    If a curve \(C\) is given in parametric form by the equations

    \[{x = x\left( t \right),\;\;}\kern0pt{y = y\left( t \right),}\]

    where the parameter \(t\) runs between \({t_1}\) and \({t_2},\) the arc length of the curve is

    \[L = \int\limits_{{t_1}}^{{t_2}} {\sqrt {{{\left[ {x’\left( t \right)} \right]}^2} + {{\left[ {y’\left( t \right)} \right]}^2}} dt} .\]

    Arc Length in Polar Coordinates

    The arc length of a polar curve given by the equation \(r = r\left( \theta \right),\) with \(\theta\) ranging over some interval \(\left[ {\alpha ,\beta } \right],\) is expressed by the formula

    \[L = \int\limits_\alpha ^\beta {\sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {{\left[ {r^\prime\left( \theta \right)} \right]}^2}} d\theta } .\]


  • Solved Problems

    Click a problem to see the solution.

    Example 1

    Find the length of the line segment given by the equation \(y = 7x + 2\) from \(x = 2\) to \(x = 6.\)

    Example 2

    Find the arc length of the semicubical parabola \(y = {x^{\frac{3}{2}}}\) from \(x = 0\) to \(x = 5.\)

    Example 3

    Prove that the circumference of a circle of radius \(R\) is \(2\pi R.\)

    Example 4

    Calculate the arc length of the curve \(y = \ln x\) from \(x = \sqrt{3}\) to \(x = \sqrt{15}.\)

    Example 5

    Find the arc length of the curve \(y = \ln \sec x\) from \(x = 0\) to \(x = \large{\frac{\pi }{3}}\normalsize.\)

    Example 6

    Find the arc length of the curve \(y = {e^x}\) from \(x = 0\) to \(x = 1.\)

    Example 7

    Find the length of one arc of the cycloid given in parametric form by the equations \(x\left( t \right) = t – \sin t,\) \(y\left( t \right) = 1 – \cos t.\)

    Example 8

    Find the arc length of the astroid \(x\left( t \right) = {\cos ^3}t,\) \(y\left( t \right) = {\sin^3}t.\)

    Example 9

    Find the length of the cardioid \(r = 1 + \cos \theta .\)

    Example 10

    Find the length of the first turn of Archimedean spiral \(r\left( \theta \right) = \theta.\)

    Example 1.

    Find the length of the line segment given by the equation \(y = 7x + 2\) from \(x = 2\) to \(x = 6.\)

    Solution.

    Let’s solve the more general problem. Consider an arbitrary straight line that is defined by the equation \(y = mx+n.\) What is the length of the line segment in the interval \(\left[ {a,b} \right]?\)

    It’s obvious that

    \[{y^\prime = f^\prime\left( x \right) }={ \left( {mx + n} \right)^\prime }={ m.}\]

    Using the arc length formula, we have

    \[{L = \int\limits_a^b {\sqrt {1 + {{\left[ {f’\left( x \right)} \right]}^2}} dx} }={ \int\limits_a^b {\sqrt {1 + {m^2}} dx} }={ \sqrt {1 + {m^2}} \left( {b – a} \right).}\]

    So, for the line segment \(y = 7x + 2,\) we obtain

    \[{L = \sqrt {1 + {7^2}} \left( {6 – 2} \right) }={ 4\sqrt {50} }={ 20\sqrt 2 .}\]

    Example 2.

    Find the arc length of the semicubical parabola \(y = {x^{\frac{3}{2}}}\) from \(x = 0\) to \(x = 5.\)

    Solution.

    We use the arc length formula

    \[L = \int\limits_a^b {\sqrt {1 + {{\left[ {f’\left( x \right)} \right]}^2}} dx} .\]

    Here \(f\left( x \right) = {x^{\frac{3}{2}}},\) \(f^\prime\left( x \right) = \left( {{x^{\frac{3}{2}}}} \right)^\prime = \large{\frac{3}{2}}\normalsize{x^{\frac{1}{2}}},\) \(a = 0,\) \(b = 5.\) Then

    \[{L = \int\limits_0^5 {\sqrt {1 + {{\left( {\frac{3}{2}{x^{\frac{1}{2}}}} \right)}^2}} dx} }={ \int\limits_0^5 {\sqrt {1 + \frac{{9x}}{4}} dx} }={ \frac{1}{2}\int\limits_0^5 {\sqrt {4 + 9x} dx} .}\]

    We make the substitution

    \[{{u^2} = 4 + 9x,}\;\; \Rightarrow {u = \sqrt {4 + 9x} ,}\;\; \Rightarrow {2udu = 9dx,}\;\; \Rightarrow {dx = \frac{{2udu}}{9}.}\]

    When \(x = 0,\) \(u = 2,\) and when \(x = 5,\) \(u = 7.\) So the integral becomes

    \[{L = \frac{1}{2}\int\limits_2^7 {\left( {u \cdot \frac{2}{9}u} \right)du} }={ \frac{1}{9}\int\limits_2^7 {{u^2}du} }={ \frac{1}{9} \cdot \left. {\frac{{{u^3}}}{3}} \right|_2^7 }={ \frac{1}{{27}}\left( {{7^3} – {2^3}} \right) }={ \frac{{335}}{{27}}.}\]

    Example 3.

    Prove that the circumference of a circle of radius \(R\) is \(2\pi R.\)

    Solution.

    Deriving the formula for the circumference of a circle
    Figure 3.

    We calculate the circumference of the upper half of the circle and then multiply the answer by \(2.\) The upper half of the circle is defined by the function

    \[y = \sqrt {{R^2} – {x^2}}. \]

    Take the derivative:

    \[\require{cancel}{y^\prime = f^\prime\left( x \right) }={ \left( {\sqrt {{R^2} – {x^2}} } \right)^\prime }={ – \frac{{\cancel{2}x}}{{\cancel{2}\sqrt {{R^2} – {x^2}} }} }={ – \frac{x}{{\sqrt {{R^2} – {x^2}} }}.}\]

    Then the circumference of the circle is given by

    \[{L = 2\int\limits_{ – R}^R {\sqrt {1 + {{\left[ {f’\left( x \right)} \right]}^2}} dx} }={ 2\int\limits_{ – R}^R {\sqrt {1 + {{\left( { – \frac{x}{{\sqrt {{R^2} – {x^2}} }}} \right)}^2}} dx} }={ 2\int\limits_{ – R}^R {\sqrt {1 + \frac{{{x^2}}}{{{R^2} – {x^2}}}} dx} }={ 2\int\limits_{ – R}^R {\sqrt {\frac{{{R^2}}}{{{R^2} – {x^2}}}} dx} }={ 2R\int\limits_{ – R}^R {\frac{{dx}}{{\sqrt {{R^2} – {x^2}} }}} }={ 2R\left. {\arcsin \frac{x}{R}} \right|_{ – R}^R }={ 2R\left[ {\arcsin 1 – \arcsin \left( { – 1} \right)} \right] }={ 2R\left[ {\frac{\pi }{2} – \left( { – \frac{\pi }{2}} \right)} \right] }={ 2\pi R.}\]

    Example 4.

    Calculate the arc length of the curve \(y = \ln x\) from \(x = \sqrt{3}\) to \(x = \sqrt{15}.\)

    Solution.

    Arc length of the natural logarithm function
    Figure 4.

    Since \(y^\prime = \left( {\ln x} \right)^\prime = \large{\frac{1}{x}}\normalsize,\) we can write

    \[{L = \int\limits_a^b {\sqrt {1 + {{\left[ {f’\left( x \right)} \right]}^2}} dx} }={ \int\limits_{\sqrt 3 }^{\sqrt {15} } {\sqrt {1 + {{\left( {\frac{1}{x}} \right)}^2}} dx} }={ \int\limits_{\sqrt 3 }^{\sqrt {15} } {\sqrt {\frac{{{x^2} + 1}}{{{x^2}}}} dx} }={ \int\limits_{\sqrt 3 }^{\sqrt {15} } {\frac{{\sqrt {{x^2} + 1} }}{x}dx} .}\]

    To evaluate the latter integral we rewrite it in the form

    \[{I = \int {\frac{{\sqrt {{x^2} + 1} }}{x}dx} }={ \int {\frac{{x\sqrt {{x^2} + 1} }}{{{x^2}}}dx} }\]

    and change the variable:

    \[{{x^2} + 1 = {u^2},}\;\; \Rightarrow {xdx = udu,}\;\; \Rightarrow {{x^2} = {u^2} – 1.}\]

    This results in

    \[{I = \int {\frac{{{u^2}}}{{{u^2} – 1}}du} }={ \int {\frac{{{u^2} – 1 + 1}}{{{u^2} – 1}}du} }={ \int {\left( {1 + \frac{1}{{{u^2} – 1}}} \right)du} }={ u + \frac{1}{2}\ln \left| {\frac{{u – 1}}{{u + 1}}} \right| }={ \sqrt {{x^2} + 1} + \frac{1}{2}\ln \left| {\frac{{\sqrt {{x^2} + 1} – 1}}{{\sqrt {{x^2} + 1} + 1}}} \right|.}\]

    Returning back to the definite integral, we find the arc length \(L:\)

    \[{L \text { = }}\kern0pt{\left. {\left[ {\sqrt {{x^2} + 1} + \frac{1}{2}\ln \left| {\frac{{\sqrt {{x^2} + 1} – 1}}{{\sqrt {{x^2} + 1} + 1}}} \right|} \right]} \right|_{\sqrt 3 }^{\sqrt {15} } }={ \left( {4 + \frac{1}{2}\ln \frac{{4 – 1}}{{4 + 1}}} \right) – \left( {2 + \frac{1}{2}\ln \frac{{2 – 1}}{{2 + 1}}} \right) }={ 2 + \frac{1}{2}\left( {\ln \frac{3}{5} }-{ \ln \frac{1}{3}} \right) }={ 2 + \frac{1}{2}\ln \frac{9}{5}.}\]

    Example 5.

    Find the arc length of the curve \(y = \ln \sec x\) from \(x = 0\) to \(x = \large{\frac{\pi }{3}}\normalsize.\)

    Solution.

    Arc length of the curve y=ln(sec x)
    Figure 5.

    The derivative of the given function is

    \[{y^\prime = f^\prime\left( x \right) }={ \left( {\ln \sec x} \right)^\prime }={ \frac{1}{{\sec x}}\left( {\sec x} \right)^\prime }={ \frac{{\left( { – {{\sec }^2}x} \right)\left( { – \sin x} \right)}}{{\sec x}} }={ \sec x\sin x }={ \tan x.}\]

    The arc length is determined by the integral

    \[{L = \int\limits_a^b {\sqrt {1 + {{\left[ {f’\left( x \right)} \right]}^2}} dx} }={ \int\limits_0^{\frac{\pi }{3}} {\sqrt {1 + {{\tan }^2}x} dx} }={ \int\limits_0^{\frac{\pi }{3}} {\sqrt {{{\sec }^2}x} dx} }={ \int\limits_0^{\frac{\pi }{3}} {\sec xdx} .}\]

    Since

    \[\int {\sec xdx} = \ln \left| {\sec x + \tan x} \right| + C,\]

    we get the following answer:

    \[{L \text{ = }}\kern0pt{\left. {\left[ {\ln \left| {\sec x + \tan x} \right|} \right]} \right|_0^{\frac{\pi }{3}} }={ \ln \left| {\sec \frac{\pi }{3} + \tan \frac{\pi }{3}} \right| }-{ \ln \left| {\sec 0 + \tan 0} \right| }={ \ln \left( {2 + \sqrt 3 } \right) – \underbrace {\ln \left( {1 + 0} \right)}_0 }={ \ln \left( {2 + \sqrt 3 } \right).}\]

    Example 6.

    Find the arc length of the exponential curve \(y = {e^x}\) from \(x = 0\) to \(x = 1.\)

    Solution.

    Arc length of the exponential curve y=exp(x) from x=0 to x=1.
    Figure 6.

    We compute the arc length using by integration:

    \[{L = \int\limits_0^1 {\sqrt {1 + {{\left[ {y^\prime\left( x \right)} \right]}^2}} dx} }={ \int\limits_0^1 {\sqrt {1 + {e^{2x}}} dx} .}\]

    To evaluate the integral, we make the following substitution:

    \[{1 + {e^{2x}} = {u^2},}\;\; \Rightarrow {{e^{2x}}dx = udu,}\;\; \Rightarrow {dx = \frac{{udu}}{{{e^{2x}}}} = \frac{{udu}}{{{u^2} – 1}}.}\]

    The indefinite integral is given by

    \[{I = \int {\frac{{{u^2}du}}{{{u^2} – 1}}} }={ \int {\frac{{{u^2} – 1 + 1}}{{{u^2} – 1}}du} }={ \int {\left( {1 + \frac{1}{{{u^2} – 1}}} \right)du} }={ u + \frac{1}{2}\ln \left| {\frac{{u – 1}}{{u + 1}}} \right| }={ \sqrt {1 + {e^{2x}}} + \frac{1}{2}\ln \left| {\frac{{\sqrt {1 + {e^{2x}}} – 1}}{{\sqrt {1 + {e^{2x}}} + 1}}} \right|.}\]

    Then the definite integral is written as

    \[{L \text{ = }}\kern0pt{\left. {\left[ {\sqrt {1 + {e^{2x}}} + \frac{1}{2}\ln \left| {\frac{{\sqrt {1 + {e^{2x}}} – 1}}{{\sqrt {1 + {e^{2x}}} + 1}}} \right|} \right]} \right|_0^1 }={ \left( {\sqrt {1 + {e^2}} + \frac{1}{2}\ln \frac{{\sqrt {1 + {e^2}} – 1}}{{\sqrt {1 + {e^2}} + 1}}} \right) }-{ \left( {\sqrt 2 + \frac{1}{2}\ln \frac{{\sqrt 2 – 1}}{{\sqrt 2 + 1}}} \right).}\]

    We simplify the fractions under the logarithm sign:

    \[{\frac{{\sqrt {1 + {e^2}} – 1}}{{\sqrt {1 + {e^2}} + 1}} }={ \frac{{{{\left( {\sqrt {1 + {e^2}} – 1} \right)}^2}}}{{{{\left( {\sqrt {1 + {e^2}} } \right)}^2} – {1^2}}} }={ \frac{{{{\left( {\sqrt {1 + {e^2}} – 1} \right)}^2}}}{{{e^2}}} }={ {\left( {\frac{{\sqrt {1 + {e^2}} – 1}}{e}} \right)^2},}\]

    \[{\frac{{\sqrt 2 – 1}}{{\sqrt 2 + 1}} }={ \frac{{{{\left( {\sqrt 2 } \right)}^2} – {1^2}}}{{{{\left( {\sqrt 2 + 1} \right)}^2}}} }={ {\left( {\frac{1}{{\sqrt 2 + 1}}} \right)^2}.}\]

    Hence, the arc length is equal to

    \[{L = \sqrt {1 + {e^2}} }-{ \sqrt 2 }+{ \frac{1}{2}\ln {\left( {\frac{{\sqrt {1 + {e^2}} – 1}}{e}} \right)^2} }-{ \frac{1}{2}\ln {\left( {\frac{1}{{\sqrt 2 + 1}}} \right)^2} }={ \sqrt {1 + {e^2}} }-{ \sqrt 2 }+{ \ln \frac{{\sqrt {1 + {e^2}} – 1}}{e} }-{ \ln \frac{1}{{\sqrt 2 + 1}} }={ \sqrt {1 + {e^2}} }-{ \sqrt 2 }+{ \ln \left[ {\left( {\sqrt {1 + {e^2}} – 1} \right)\left( {\sqrt 2 + 1} \right)} \right] }-{ \underbrace {\ln e}_1 }={ \sqrt {1 + {e^2}} }-{ \sqrt 2 – 1 }+{ \ln \left[ {\left( {\sqrt {1 + {e^2}} – 1} \right)\left( {\sqrt 2 + 1} \right)} \right]}\approx {2.004}\]

    Example 7.

    Find the length of one arc of the cycloid given in parametric form by the equations \(x\left( t \right) = t – \sin t,\) \(y\left( t \right) = 1 – \cos t.\)

    Solution.

    One arc of the cycloid
    Figure 7.

    The arc length of a parametric curve is expressed by the integral

    \[L = \int\limits_{{t_1}}^{{t_2}} {\sqrt {{{\left[ {x^\prime\left( t \right)} \right]}^2} + {{\left[ {y^\prime\left( t \right)} \right]}^2}} dt} .\]

    For the first arc of cycloid, \(0 \le t \le 2\pi .\) Hence,

    \[{L \text{ = }}\kern0pt{\int\limits_0^{2\pi } {\sqrt {{{\left( {1 – \cos t} \right)}^2} + {{\sin }^2}t} \,dt} }={ \int\limits_0^{2\pi } {\sqrt {1 – 2\cos t + \underbrace {{{\cos }^2}t + {{\sin }^2}t}_1} \,dt} }={ \int\limits_0^{2\pi } {\sqrt {2 – 2\cos t} \,dt} }={ \int\limits_0^{2\pi } {\sqrt {4{{\sin }^2}\frac{t}{2}} dt} }={ 2\int\limits_0^{2\pi } {\sin \frac{t}{2}dt} }={ 4\left. {\left( { – \cos \frac{t}{2}} \right)} \right|_0^{2\pi } }={ 4\left( { – \cos \pi + \cos 0} \right) }={ 8.}\]

    Example 8.

    Find the arc length of the astroid \(x\left( t \right) = {\cos ^3}t,\) \(y\left( t \right) = {\sin^3}t.\)

    Solution.

    Arc length of the astroid given in parametric form
    Figure 8.

    As the curve is given in parametric form, we use the formula

    \[{L \text{ = }}\kern0pt{\int\limits_a^b {\sqrt {{{\left[ {x^\prime\left( t \right)} \right]}^2} + {{\left[ {y^\prime\left( t \right)} \right]}^2}} dt} .}\]

    Find the derivatives:

    \[{x^\prime\left( t \right) = \left( {{{\cos }^3}t} \right)^\prime }={ 3{\cos ^2}t\left( { – \sin t} \right) }={ – 3{\cos ^2}t\sin t,}\]

    \[{y^\prime\left( t \right) = \left( {{{\sin }^3}t} \right)^\prime }={ 3{\sin ^2}t\cos t.}\]

    We calculate the length of one arc of the astroid lying in the first quadrant and then multiply the result by \(4.\) So, we have

    \[{L \text{ = }}\kern0pt{4\int\limits_0^{\frac{\pi }{2}} {\sqrt {9\,{{\cos }^4}t\,{{\sin }^2}t + 9\,{{\sin }^4}t\,{{\cos }^2}t} \,dt} }={ 4\int\limits_0^{\frac{\pi }{2}} {\sqrt {9\,{{\sin }^2}t\,{{\cos }^2}t\underbrace {\left( {{{\cos }^2}t + {{\sin }^2}t} \right)}_1} dt} }={ 12\int\limits_0^{\frac{\pi }{2}} {\sin t\cos tdt} }={ 6\int\limits_0^{\frac{\pi }{2}} {\sin 2tdt} }={ – 3\left. {\cos 2t} \right|_0^{\frac{\pi }{2}} }={ – 3\left( {\cos \pi – \cos 0} \right) }={ 6.}\]

    Example 9.

    Find the length of the cardioid \(r = 1 + \cos \theta .\)

    Solution.

    Arc length of the cardioid r=1+cos(theta).
    Figure 9.

    The cardioid is given in polar coordinates. Therefore we use the formula

    \[L = \int\limits_\alpha ^\beta {\sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {{\left[ {r^\prime\left( \theta \right)} \right]}^2}} d\theta } .\]

    Due to symmetry, we calculate the arc length of the upper half of the cardioid (with \(\theta\) ranging from \(0\) to \(\pi\)) and multiply the result by \(2.\) This yields

    \[{L = \int\limits_\alpha ^\beta {\sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {{\left[ {r’\left( \theta \right)} \right]}^2}} d\theta } }={ 2\int\limits_0^\pi {\sqrt {{{\left( {1 + \cos \theta } \right)}^2} + {{\left( { – \sin \theta } \right)}^2}} d\theta } }={ 2\int\limits_0^\pi {\sqrt {1 + 2\cos \theta + \underbrace {{{\cos }^2}\theta + {{\sin }^2}\theta }_1} \,d\theta } }={ 2\int\limits_0^\pi {\sqrt {2 + 2\cos \theta } \,d\theta } }={ 2\sqrt 2 \int\limits_0^\pi {\sqrt {1 + \cos \theta } \,d\theta } }={ 2\sqrt 2 \int\limits_0^\pi {\sqrt {2{{\cos }^2}\frac{\theta }{2}} d\theta } }={ 4\int\limits_0^\pi {\cos \frac{\theta }{2}d\theta } }={ 8\left. {\sin \frac{\theta }{2}} \right|_0^\pi }={ 8\left( {\sin \frac{\pi }{2} – \sin 0} \right) }={ 8.}\]

    Example 10.

    Find the length of the first turn of Archimedean spiral \(r\left( \theta \right) = \theta.\)

    Solution.

    The length of the first turn of Archimedean spiral r=theta.
    Figure 10.

    The arc length of a curve in polar coordinates is given by the equation

    \[L = \int\limits_\alpha ^\beta {\sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {{\left[ {r^\prime\left( \theta \right)} \right]}^2}} d\theta } .\]

    Here we have \(\alpha = 0,\) \(\beta = 2\pi,\) \(r\left( \theta \right) = \theta,\) \(r^\prime\left( \theta \right) = 1,\) so the arc length is expressed by the integral

    \[L = \int\limits_0^{2\pi } {\sqrt {1 + {\theta ^2}} d\theta } .\]

    We use the trig substitution \(\theta = \tan t\) to evaluate the integral. Substituting

    \[{\theta = \tan t,\;\;}\kern0pt{\sqrt {1 + {\theta ^2}} = \sec t,\;\;}\kern0pt{d\theta = {\sec ^2}t\,dt,}\]

    we rewrite the indefinite integral in the form

    \[{I = \int {\sqrt {1 + {\theta ^2}} d\theta } }={ \int {{{\sec }^3}tdt} .}\]

    Now we use the reduction formula

    \[{\int {{{\sec }^3}tdt} }={ \frac{{\sec t\tan t}}{2} }+{ \frac{1}{2}\int {\sec tdt} .}\]

    Recall that the integral of secant is common and is given by

    \[\int {\sec tdt} = \ln \left| {\sec t + \tan t} \right|.\]

    Then

    \[{\int {{{\sec }^3}tdt} }={ \frac{{\sec t\tan t}}{2} }+{ \frac{1}{2}\ln \left| {\sec t + \tan t} \right|.}\]

    Returning back to the variable \(x,\) we obtain:

    \[{I = \frac{{\theta \sqrt {1 + {\theta ^2}} }}{2} }+{ \frac{1}{2}\ln \left| {\sqrt {1 + {\theta ^2}} + \theta } \right|.}\]

    Hence, the arc length is

    \[{L \text{ = }}\kern0pt{\left. {\left[ {\frac{{\theta \sqrt {1 + {\theta ^2}} }}{2} + \frac{1}{2}\ln \left| {\sqrt {1 + {\theta ^2}} + \theta } \right|} \right]} \right|_0^{2\pi } }={ \left( {\frac{{\cancel{2}\pi \sqrt {1 + 4{\pi ^2}} }}{\cancel{2}} }\right.}+{\left.{ \frac{1}{2}\ln \left| {\sqrt {1 + 4{\pi ^2}} + 2\pi } \right|} \right) }-{ \underbrace {\left( {0 + \frac{1}{2}\ln 1} \right)}_0 }={ \pi \sqrt {1 + 4{\pi ^2}} + \frac{1}{2}\ln \left( {\sqrt {1 + 4{\pi ^2}} + 2\pi } \right)} \approx {21.26}\]