What is the integral from-1 to x of sin(t^2) ?

The integral from-1 to x of sin(t^2) is sqrt(pi/2)\S(sqrt(2/pi)x)-sqrt(pi/2)\S(-1*sqrt(2/pi))

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

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

\frac{π}{2} § (\frac{2}{π} x) - \frac{π}{2} § (- 1 \cdot \frac{2}{π})

Solution steps

\int_{- 1}^{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)]_{- 1}^{x}

Compute the boundaries:

\frac{π}{2} § (\frac{2}{π} x) - \frac{π}{2} § (- 1 \cdot \frac{2}{π})

= \frac{π}{2} § (\frac{2}{π} x) - \frac{π}{2} § (- 1 \cdot \frac{2}{π})

\int_{- 1}^{1} 5 x^{2} - 5 d x

\int_{1}^{12} \frac{4}{x ^{2}} d x

\int_{- 1}^{2} x^{3} + 1 d x

\int_{0}^{2} 1 - x^{3} d x

\int_{- \frac{1}{2}}^{\frac{1}{2}} cos (π x) d x

What is the integral from-1 to x of sin(t^2) ?
The integral from-1 to x of sin(t^2) is sqrt(pi/2)\S(sqrt(2/pi)x)-sqrt(pi/2)\S(-1*sqrt(2/pi))