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

The integral from pi to 0 of (sin(x)-3)^2 is -pi/2+12-9pi

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

\int_{π}^{0} (sin (x) - 3)^{2} d x

- \frac{π}{2} + 12 - 9 π

Decimal

- 17.84513 \dots

Solution steps

\int_{π}^{0} (sin (x) - 3)^{2} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Factor

sin (x) - 3 : - (- sin (x) + 3)

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

Simplify

(- (- sin (x) + 3))^{2} : (- sin (x) + 3)^{2}

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

Expand

(- sin (x) + 3)^{2} : sin^{2} (x) - 6 sin (x) + 9

= - \int_{0}^{π} sin^{2} (x) - 6 sin (x) + 9 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}^{π} sin^{2} (x) d x - \int_{0}^{π} 6 sin (x) d x + \int_{0}^{π} 9 d x)

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

\int_{0}^{π} 6 sin (x) d x = 12

\int_{0}^{π} 9 d x = 9 π

= - (\frac{π}{2} - 12 + 9 π)

Simplify

= - \frac{π}{2} + 12 - 9 π

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What is the integral from pi to 0 of (sin(x)-3)^2 ?
The integral from pi to 0 of (sin(x)-3)^2 is -pi/2+12-9pi