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人気のある微分積分 >

integral from 0 to pi/4 of sin^4(2x)

解

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

解

\frac{3 π}{32}

+1

十進法表記

0.29452 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{0}^{\frac{π}{2}} sin^{4} (u) \frac{1}{2} d u

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}} sin^{4} (u) d u

積分換算を適用する

= \frac{1}{2} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) d u)

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

= \frac{1}{2} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \frac{π}{4})

簡素化

\frac{1}{2} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \frac{π}{4}) : \frac{1}{2} (- [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{\frac{π}{2}} + \frac{3 π}{16})

= \frac{1}{2} (- [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{\frac{π}{2}} + \frac{3 π}{16})

限度を計算する:

0

= \frac{1}{2} (- 0 + \frac{3 π}{16})

簡素化

= \frac{3 π}{32}

グラフ

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