integral from sqrt(x) to x of sin(t^2-1)

解答

\int_{x}^{x} sin (t^{2} - 1) d t

- sin (1) (\frac{π}{2} C (\frac{2}{π} x) - \frac{π}{2} C (\frac{2}{π} x)) + cos (1) (\frac{π}{2} § (\frac{2}{π} x) - \frac{π}{2} § (\frac{2}{π} x))

求解步骤

\int_{x}^{x} sin (t^{2} - 1) d t

使用三角恒等式改写

= \int_{x}^{x} - cos (t^{2}) sin (1) + cos (1) sin (t^{2}) 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

= - \int_{x}^{x} cos (t^{2}) sin (1) d t + \int_{x}^{x} cos (1) sin (t^{2}) d t

\int_{x}^{x} cos (t^{2}) sin (1) d t = sin (1) (\frac{π}{2} C (\frac{2}{π} x) - \frac{π}{2} C (\frac{2}{π} x))

\int_{x}^{x} cos (1) sin (t^{2}) d t = cos (1) (\frac{π}{2} § (\frac{2}{π} x) - \frac{π}{2} § (\frac{2}{π} x))

= - sin (1) (\frac{π}{2} C (\frac{2}{π} x) - \frac{π}{2} C (\frac{2}{π} x)) + cos (1) (\frac{π}{2} § (\frac{2}{π} x) - \frac{π}{2} § (\frac{2}{π} x))

\int_{2}^{4} - x + 4 d x

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