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

解答

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

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

求解步骤

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

使用三角恒等式改写

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

提出常数:

\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

使用换元积分法

= \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)

提出常数:

\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))

乘开

\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))

使用积分加法定则:

\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})))

化简

\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

\int_{1}^{b} \frac{1}{x ^{p}} d x

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\int_{2}^{6} (- x^{2} + 8 x - 12) d x

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\int_{0}^{1} (- 1 + \frac{2}{1 + t}) d t