What is the integral from 4x to-5 of sin(t^2) ?

The integral from 4x to-5 of sin(t^2) is sqrt(pi/2)\S(-5sqrt(2/pi))-sqrt(pi/2)\S(sqrt(2/pi)*4x)

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integral from 4x to-5 of sin(t^2)

\int_{4 x}^{- 5} sin (t^{2}) d t

\frac{π}{2} § (- 5 \frac{2}{π}) - \frac{π}{2} § (\frac{2}{π} \cdot 4 x)

Solution steps

\int_{4 x}^{- 5} 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)]_{4 x}^{- 5}

Compute the boundaries:

\frac{π}{2} § (- 5 \frac{2}{π}) - \frac{π}{2} § (\frac{2}{π} \cdot 4 x)

= \frac{π}{2} § (- 5 \frac{2}{π}) - \frac{π}{2} § (\frac{2}{π} \cdot 4 x)

What is the integral from 4x to-5 of sin(t^2) ?
The integral from 4x to-5 of sin(t^2) is sqrt(pi/2)\S(-5sqrt(2/pi))-sqrt(pi/2)\S(sqrt(2/pi)*4x)