integral from 0 to x of cosh^2(t)

解答

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

\frac{1}{2} (x + \frac{1}{2} sinh (2 x))

求解步骤

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

使用三角恒等式改写

= \int_{0}^{x} \frac{1 + cosh ( 2 t )}{2} d t

提出常数:

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

= \frac{1}{2} \cdot \int_{0}^{x} 1 + cosh (2 t) d t

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}{2} (\int_{0}^{x} 1 d t + \int_{0}^{x} cosh (2 t) d t)

\int_{0}^{x} 1 d t = x

\int_{0}^{x} cosh (2 t) d t = \frac{1}{2} sinh (2 x)

= \frac{1}{2} (x + \frac{1}{2} sinh (2 x))

\int_{0}^{x} \frac{1}{1 + t + t ^{2}} d t

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