What is the integral from 0 to 2x of sin(t^2) ?

The integral from 0 to 2x of sin(t^2) is sqrt(pi/2)\S(sqrt(2/pi)*2x)-sqrt(pi/2)\S(sqrt(2/pi)*0)

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

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

\frac{π}{2} § (\frac{2}{π} \cdot 2 x) - \frac{π}{2} § (\frac{2}{π} \cdot 0)

Solution steps

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

Use the nonelementary integral:

\int sin (x^{2}) d x = \frac{π}{2} § (\frac{2}{π} x)

= [\frac{π}{2} § (\frac{2}{π} t)]_{0}^{2 x}

Compute the boundaries:

\frac{π}{2} § (\frac{2}{π} \cdot 2 x) - \frac{π}{2} § (\frac{2}{π} \cdot 0)

= \frac{π}{2} § (\frac{2}{π} \cdot 2 x) - \frac{π}{2} § (\frac{2}{π} \cdot 0)

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What is the integral from 0 to 2x of sin(t^2) ?
The integral from 0 to 2x of sin(t^2) is sqrt(pi/2)\S(sqrt(2/pi)*2x)-sqrt(pi/2)\S(sqrt(2/pi)*0)