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

The integral from 0 to pi/2 of (2-sin(x))^2 is 2pi-4+pi/4

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

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

2 π - 4 + \frac{π}{4}

Decimal

3.06858 \dots

Solution steps

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

Expand

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

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

\int_{0}^{\frac{π}{2}} 4 d x = 2 π

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

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

= 2 π - 4 + \frac{π}{4}

\int_{\frac{2}{3}}^{2} t arcsec (t) d t

\int_{0}^{π} (1 - 2 x + e^{x}) d x

\int_{1}^{6} (8 t^{3} - 6 t^{- 2}) d t

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

\int_{- π}^{π} 2 x d x

What is the integral from 0 to pi/2 of (2-sin(x))^2 ?
The integral from 0 to pi/2 of (2-sin(x))^2 is 2pi-4+pi/4