What is area sin(x),[ 1/9 pi, 1/8 pi] ?

The answer to area sin(x),[ 1/9 pi, 1/8 pi] is -(sqrt(2+\sqrt{2)})/2+cos(pi/9)

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area sin(x),[ 1/9 pi, 1/8 pi]

area

sin (x), [\frac{1}{9} π, \frac{1}{8} π]

- \frac{2 + 2}{2} + cos (\frac{π}{9})

Decimal

0.01581 \dots

Solution steps

Apply the area formula:

\int_{\frac{1}{9} π}^{\frac{1}{8} π} ∣ sin (x) ∣ d x

Solve

\int_{\frac{1}{9} π}^{\frac{1}{8} π} ∣ sin (x) ∣ d x : - \frac{2 + 2}{2} + cos (\frac{π}{9})

The area is:

= - \frac{2 + 2}{2} + cos (\frac{π}{9})

\int sin (2 a) x d x

\frac{d}{d x} (\frac{x + 5}{x ^{2} + x - 6})

f (x) = x^{\frac{1}{5}} + 8

(- 2, 0), (0, - 4)

x \to 3 lim (h (x) - g (x))

What is area sin(x),[ 1/9 pi, 1/8 pi] ?
The answer to area sin(x),[ 1/9 pi, 1/8 pi] is -(sqrt(2+\sqrt{2)})/2+cos(pi/9)