What is area y=6cos(3x),y=6sin(3x),(0, pi/8) ?

The answer to area y=6cos(3x),y=6sin(3x),(0, pi/8) is -sqrt(2-\sqrt{2)}-sqrt(2+\sqrt{2)}+4sqrt(2)-2

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area y=6cos(3x),y=6sin(3x),(0, pi/8)

area

y = 6 cos (3 x), y = 6 sin (3 x), (0, \frac{π}{8})

- 2 - 2 - 2 + 2 + 42 - 2

Decimal Notation

1.04372 \dots

Solution steps

Apply the area formula:

\int_{0}^{\frac{π}{8}} ∣ 6 cos (3 x) - 6 sin (3 x) ∣ d x

Solve

\int_{0}^{\frac{π}{8}} ∣ 6 cos (3 x) - 6 sin (3 x) ∣ d x : - 2 - 2 - 2 + 2 + 42 - 2

The area is:

= - 2 - 2 - 2 + 2 + 42 - 2

\frac{\partial}{\partial x} (\frac{x}{x ^{5} - y ^{5}})

x \to 2 lim (\frac{∣ 2 - x ∣}{2 - x})

\int sin (x) csc (x) d x

f (x) = e^{2 x^{2}}

\frac{1}{s ^{2} ( s ^{2} + 4 s + 5 )}

What is area y=6cos(3x),y=6sin(3x),(0, pi/8) ?
The answer to area y=6cos(3x),y=6sin(3x),(0, pi/8) is -sqrt(2-\sqrt{2)}-sqrt(2+\sqrt{2)}+4sqrt(2)-2