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人気のある微分積分 >

integral from 0 to pi/(12) of xtan^2(4x)

解

\int_{0}^{\frac{π}{12}} x tan^{2} (4 x) d x

解

\frac{- 3 π ^{2} + 18 π - 18 3 ln ( 2 )}{288 3}

+1

十進法表記

0.03577 \dots

解答ステップ

\int_{0}^{\frac{π}{12}} x tan^{2} (4 x) d x

三角関数の公式を使用して書き換える

= \int_{0}^{\frac{π}{12}} x (- 1 + sec^{2} (4 x)) d x

u置換積分法を適用する

= \int_{0}^{\frac{π}{3}} \frac{u tan ^{2} ( u )}{16} d u

定数を除く:

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

= \frac{1}{16} \cdot \int_{0}^{\frac{π}{3}} u tan^{2} (u) d u

部分積分を適用する

= \frac{1}{16} [u (- u + tan (u)) - \int - u + tan (u) d u]_{0}^{\frac{π}{3}}

\int - u + tan (u) d u = - \frac{u ^{2}}{2} - ln ∣ cos (u) ∣

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

簡素化

\frac{1}{16} [u (- u + tan (u)) - (- \frac{u ^{2}}{2} - ln ∣ cos (u) ∣)]_{0}^{\frac{π}{3}} : \frac{1}{16} [\frac{- u ^{2} + 2 u tan ( u ) + 2 ln ∣ cos ( u ) ∣}{2}]_{0}^{\frac{π}{3}}

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

限度を計算する:

\frac{- 3 π ^{2} + 18 π - 18 3 ln ( 2 )}{18 3}

= \frac{1}{16} \cdot \frac{- 3 π ^{2} + 18 π - 18 3 ln ( 2 )}{18 3}

簡素化

= \frac{- 3 π ^{2} + 18 π - 18 3 ln ( 2 )}{288 3}

グラフ

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