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

解答

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

2 π - 4 + \frac{π}{4}

十进制

3.06858 \dots

求解步骤

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

乘开

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

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

使用积分加法定则:

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

\int_{0}^{\frac{π}{2}} 4 d θ = 2 π

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

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

= 2 π - 4 + \frac{π}{4}

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