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人気のある微分積分 >

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

解

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

解

\frac{π ^{2} - 4 π + 8}{16}

+1

十進法表記

0.33145 \dots

解答ステップ

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

三角関数の公式を使用して書き換える

= \int_{0}^{\frac{π}{2}} \frac{1}{2} θ (1 - cos (θ)) d θ

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} θ (1 - cos (θ)) d θ

部分積分を適用する

= \frac{1}{2} [θ (θ - sin (θ)) - \int θ - sin (θ) d θ]_{0}^{\frac{π}{2}}

\int θ - sin (θ) d θ = \frac{θ ^{2}}{2} + cos (θ)

= \frac{1}{2} [θ (θ - sin (θ)) - (\frac{θ ^{2}}{2} + cos (θ))]_{0}^{\frac{π}{2}}

簡素化

\frac{1}{2} [θ (θ - sin (θ)) - (\frac{θ ^{2}}{2} + cos (θ))]_{0}^{\frac{π}{2}} : \frac{1}{2} [\frac{θ ^{2} - 2 θ sin ( θ ) - 2 cos ( θ )}{2}]_{0}^{\frac{π}{2}}

= \frac{1}{2} [\frac{θ ^{2} - 2 θ sin ( θ ) - 2 cos ( θ )}{2}]_{0}^{\frac{π}{2}}

限度を計算する:

\frac{π ^{2} - 4 π}{8} + 1

= \frac{1}{2} (\frac{π ^{2} - 4 π}{8} + 1)

簡素化

= \frac{π ^{2} - 4 π + 8}{16}

グラフ

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