What is the integral from 0 to pi/3 of (1-2sin(x))^2 ?

The integral from 0 to pi/3 of (1-2sin(x))^2 is (2pi-sqrt(3))/2-2

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integral from 0 to pi/3 of (1-2sin(x))^2

\int_{0}^{\frac{π}{3}} (1 - 2 sin (x))^{2} d x

\frac{2 π - 3}{2} - 2

Decimal Notation

0.27556 \dots

Solution steps

\int_{0}^{\frac{π}{3}} (1 - 2 sin (x))^{2} d x

Factor

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

= \int_{0}^{\frac{π}{3}} (- (- 1 + 2 sin (x)))^{2} d x

Simplify

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

= \int_{0}^{\frac{π}{3}} (- 1 + 2 sin (x))^{2} d x

Expand

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

= \int_{0}^{\frac{π}{3}} 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_{0}^{\frac{π}{3}} 1 d x - \int_{0}^{\frac{π}{3}} 4 sin (x) d x + \int_{0}^{\frac{π}{3}} 4 sin^{2} (x) d x

\int_{0}^{\frac{π}{3}} 1 d x = \frac{π}{3}

\int_{0}^{\frac{π}{3}} 4 sin (x) d x = 2

\int_{0}^{\frac{π}{3}} 4 sin^{2} (x) d x = 2 (\frac{π}{3} - \frac{3}{4})

= \frac{π}{3} - 2 + 2 (\frac{π}{3} - \frac{3}{4})

Simplify

\frac{π}{3} - 2 + 2 (\frac{π}{3} - \frac{3}{4}) : \frac{2 π - 3}{2} - 2

= \frac{2 π - 3}{2} - 2

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What is the integral from 0 to pi/3 of (1-2sin(x))^2 ?
The integral from 0 to pi/3 of (1-2sin(x))^2 is (2pi-sqrt(3))/2-2