integral from 0 to a of cosh^2(t/a)

解

\int_{0}^{a} cosh^{2} (\frac{t}{a}) d t

\frac{e ^{4} + 4 e ^{2} - 1}{8 e ^{2}} a

解答ステップ

\int_{0}^{a} cosh^{2} (\frac{t}{a}) d t

三角関数の公式を使用して書き換える

= \int_{0}^{a} \frac{1}{4} (e^{- \frac{t}{a}} + e^{\frac{t}{a}})^{2} d t

定数を除く:

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

= \frac{1}{4} \cdot \int_{0}^{a} (e^{- \frac{t}{a}} + e^{\frac{t}{a}})^{2} d t

u置換積分法を適用する

= \frac{1}{4} \cdot \int_{1}^{\frac{1}{e}} - \frac{a ( u ^{4} + 2 u ^{2} + 1 )}{u ^{3}} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= \frac{1}{4} (- \int_{\frac{1}{e}}^{1} - \frac{a ( u ^{4} + 2 u ^{2} + 1 )}{u ^{3}} d u)

定数を除く:

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

= \frac{1}{4} (- (- a \cdot \int_{\frac{1}{e}}^{1} \frac{u ^{4} + 2 u ^{2} + 1}{u ^{3}} d u))

拡張

\frac{u ^{4} + 2 u ^{2} + 1}{u ^{3}} : u + \frac{2}{u} + \frac{1}{u ^{3}}

= \frac{1}{4} (- (- a \cdot \int_{\frac{1}{e}}^{1} u + \frac{2}{u} + \frac{1}{u ^{3}} d u))

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (- (- a (\int_{\frac{1}{e}}^{1} u d u + \int_{\frac{1}{e}}^{1} \frac{2}{u} d u + \int_{\frac{1}{e}}^{1} \frac{1}{u ^{3}} d u)))

\int_{\frac{1}{e}}^{1} u d u = \frac{1}{2} - \frac{1}{2 e ^{2}}

\int_{\frac{1}{e}}^{1} \frac{2}{u} d u = 2

\int_{\frac{1}{e}}^{1} \frac{1}{u ^{3}} d u = \frac{- 1 + e ^{2}}{2}

= \frac{1}{4} (- (- a (\frac{1}{2} - \frac{1}{2 e ^{2}} + 2 + \frac{- 1 + e ^{2}}{2})))

簡素化

\frac{1}{4} (- (- a (\frac{1}{2} - \frac{1}{2 e ^{2}} + 2 + \frac{- 1 + e ^{2}}{2}))) : \frac{e ^{4} + 4 e ^{2} - 1}{8 e ^{2}} a

= \frac{e ^{4} + 4 e ^{2} - 1}{8 e ^{2}} a