integral from 0 to 1 of sinh(4x)(x^2+5)

解

\int_{0}^{1} sinh (4 x) (x^{2} + 5) d x

\frac{45 e ^{8} + 53}{64 e ^{4}} - \frac{41}{32}

十進法表記

37.12324 \dots

解答ステップ

\int_{0}^{1} sinh (4 x) (x^{2} + 5) d x

部分積分を適用する

= [\frac{1}{4} cosh (4 x) (x^{2} + 5) - \int \frac{1}{2} x cosh (4 x) d x]_{0}^{1}

\int \frac{1}{2} x cosh (4 x) d x = \frac{1}{2} (\frac{1}{4} x sinh (4 x) - \frac{1}{16} cosh (4 x))

= [\frac{1}{4} cosh (4 x) (x^{2} + 5) - \frac{1}{2} (\frac{1}{4} x sinh (4 x) - \frac{1}{16} cosh (4 x))]_{0}^{1}

簡素化

[\frac{1}{4} cosh (4 x) (x^{2} + 5) - \frac{1}{2} (\frac{1}{4} x sinh (4 x) - \frac{1}{16} cosh (4 x))]_{0}^{1} : [\frac{1}{32} (8 x^{2} cosh (4 x) - 4 x sinh (4 x) + 41 cosh (4 x))]_{0}^{1}

= [\frac{1}{32} (8 x^{2} cosh (4 x) - 4 x sinh (4 x) + 41 cosh (4 x))]_{0}^{1}

限度を計算する:

\frac{45 e ^{8} + 53}{64 e ^{4}} - \frac{41}{32}

= \frac{45 e ^{8} + 53}{64 e ^{4}} - \frac{41}{32}