Length of a Curve

For a parametric function x = f(t) and y = g(t),
the length of a curve from t = a to t = b is
l = ∫_a^b √(dx/dt)² + (dy/dt)² dt.

x = t - sin t

So dx/dt = 1 - cos t.

Derivative of a Polynomial

Derivative of sin x

y = 1 - cos t

So dy/dt = sin t.

Derivative of cos x

t is from 0 to 2π.
dx/dt = 1 - cos t
dy/dt = sin t

So l = ∫₀^2π √(1 - cos t)² + sin² t dt.

(1 - cos t)² = 1 - 2 cos t + cos² t

Square of a Difference: (a - b)²

cos² t + sin² t = 1

Pythagorean Identity

1 - 2 cos t + 1 = 2 - 2 cos t

2 - 2 cos t = 2(1 - cos t)

2(1 - cos t) = 4⋅[(1 - cos t)/2]

4 = 2²

(1 - cos t)/2 = sin² t/2

sin A/2

√2²⋅[sin² t/2] = |2 sin t/2|

Simplify a Radical

You should check
if 2 sin t/2 ≥ 0
from t = 0 to t = 2π.
So write absolue value signs.

If t = 0,
2 sin t/2 = 2 sin 0/2 = 2⋅0 = 0.

If t = 2π,
2 sin t/2 = 2 sin 2π/2 = 2 sin π = 2⋅0 = 0.

And any t between 0 and 2π
makes 2 sin t/2 plus.

So you can remove the absolue value signs.

Sine: Graph

Take the constant 2
out from the integral.

Solve the integral.

Definite Integral: How to Solve

The integral of sin t/2 is
2⋅(-cos t/2).

Linear Change of Variable Rule

Integral of sin x

2[2⋅(-cos t/2)]₀^2π = -4[cos t/2]₀^2π

Put 0 and 2π
into cos t/2.

cos 2π/2 = cos π
-cos 0/2 = -cos 0

cos π = (-1)
-cos 0 = -1

Cosine Values of Commonly Used Angles

(-1) - 1 = -2

-4⋅(-2) = 8

So 8 is the answer.

For a function y = f(x),
the length of a curve from x = a to x = b is
l = ∫_a^b √1 + [y']² dx.

y = [1/2]x² - [1/4][ln x]

Then y' = x - 1/4x.

Derivative of ln x

x is from 1 to e.
y' = x - 1/4x

Then l = ∫₁^e √1 + (x - 1/4x)² dx.

(x - 1/4x)² = x² - 2⋅x⋅[1/4x] + [1/4x]²

-2⋅x⋅[1/4x] = -1/2

1 - 1/2 = +1/2

You've changed -2⋅x⋅[1/4x] to -1/2.

So, change +1/2 to +2⋅x⋅[1/4x].

x² + 2⋅x⋅[1/4x] + [1/4x]² = (x + 1/4x)²

Factor a Perfect Square Trinomial

√(x + 1/4x)² = |x + 1/4x|

Again, you should check
if x + 1/4x ≥ 0
from x = 1 to x = e.
So write absolue value signs.

Any number from x = 1 to x = e
satisfies x + 1/4x ≥ 0.

So you can remove the absolute value signs.

Solve the integral.

The integral of x is
[1/2]x².

Integral of a Polynomial

The integral of +1/4x is
+[1/4] ln |x|.

Integral of 1/x

Put e and 1
into x + [1/4] ln |x|.

[1/2]⋅e² = e²/2
+[1/4] ln e = +[1/4]⋅1

[1/2]⋅1² = 1/2
+[1/4] ln e = +[1/4]⋅0

+[1/4]⋅1 = +1/4

+[1/4]⋅0 = +0

-(1/2 + 0) = -1/2

To solve +1/4 - 1/2,
multiply 2/2 to -1/2.
Then -1/2 = -[1/2]⋅[2/2] = -2/4.

+1/4 - 2/4 = -1/4

So e²/2 - 1/4 is the answer.