What is the integral from pi/6 to (5pi)/6 of (1-2sin(x))^2 ?

The integral from pi/6 to (5pi)/6 of (1-2sin(x))^2 is -3sqrt(3)+2pi

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integral from pi/6 to (5pi)/6 of (1-2sin(x))^2

\int_{\frac{π}{6}}^{\frac{5 π}{6}} (1 - 2 sin (x))^{2} d x

- 33 + 2 π

Decimal

1.08703 \dots

Solution steps

\int_{\frac{π}{6}}^{\frac{5 π}{6}} (1 - 2 sin (x))^{2} d x

Factor

1 - 2 sin (x) : - (- 1 + 2 sin (x))

= \int_{\frac{π}{6}}^{\frac{5 π}{6}} (- (- 1 + 2 sin (x)))^{2} d x

Simplify

(- (- 1 + 2 sin (x)))^{2} : (- 1 + 2 sin (x))^{2}

= \int_{\frac{π}{6}}^{\frac{5 π}{6}} (- 1 + 2 sin (x))^{2} d x

Expand

(- 1 + 2 sin (x))^{2} : 1 - 4 sin (x) + 4 sin^{2} (x)

= \int_{\frac{π}{6}}^{\frac{5 π}{6}} 1 - 4 sin (x) + 4 sin^{2} (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{\frac{π}{6}}^{\frac{5 π}{6}} 1 d x - \int_{\frac{π}{6}}^{\frac{5 π}{6}} 4 sin (x) d x + \int_{\frac{π}{6}}^{\frac{5 π}{6}} 4 sin^{2} (x) d x

\int_{\frac{π}{6}}^{\frac{5 π}{6}} 1 d x = \frac{2 π}{3}

\int_{\frac{π}{6}}^{\frac{5 π}{6}} 4 sin (x) d x = 43

\int_{\frac{π}{6}}^{\frac{5 π}{6}} 4 sin^{2} (x) d x = \frac{4 π}{3} + 3

= \frac{2 π}{3} - 43 + \frac{4 π}{3} + 3

Simplify

\frac{2 π}{3} - 43 + \frac{4 π}{3} + 3 : - 33 + 2 π

= - 33 + 2 π

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What is the integral from pi/6 to (5pi)/6 of (1-2sin(x))^2 ?
The integral from pi/6 to (5pi)/6 of (1-2sin(x))^2 is -3sqrt(3)+2pi