What is area 2cos(9x),2-2cos(9x),[0, pi/9 ] ?

The answer to area 2cos(9x),2-2cos(9x),[0, pi/9 ] is 4/(3sqrt(3))+(2pi)/(27)

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area 2cos(9x),2-2cos(9x),[0, pi/9 ]

area

2 cos (9 x), 2 - 2 cos (9 x), [0, \frac{π}{9}]

\frac{4}{3 3} + \frac{2 π}{27}

Decimal Notation

1.00251 \dots

Solution steps

Apply the area formula:

\int_{0}^{\frac{π}{9}} ∣ 2 cos (9 x) - (2 - 2 cos (9 x)) ∣ d x

Solve

\int_{0}^{\frac{π}{9}} ∣ 2 cos (9 x) - (2 - 2 cos (9 x)) ∣ d x : \frac{4}{3 3} + \frac{2 π}{27}

The area is:

= \frac{4}{3 3} + \frac{2 π}{27}

\int 6 arctan (x) d x

x \to 8 + lim (\frac{1}{∣ 8 - x ∣})

sin (8 x)

\frac{d y}{d x} = b - a y

y^{^{'}} = x (1 + y^{2})

What is area 2cos(9x),2-2cos(9x),[0, pi/9 ] ?
The answer to area 2cos(9x),2-2cos(9x),[0, pi/9 ] is 4/(3sqrt(3))+(2pi)/(27)