What is the integral from 7x to-8 of sin(t^2) ?

The integral from 7x to-8 of sin(t^2) is sqrt(pi/2)\S(-8sqrt(2/pi))-sqrt(pi/2)\S(sqrt(2/pi)*7x)

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integral from 7x to-8 of sin(t^2)

\int_{7 x}^{- 8} sin (t^{2}) d t

\frac{π}{2} § (- 8 \frac{2}{π}) - \frac{π}{2} § (\frac{2}{π} \cdot 7 x)

Solution steps

\int_{7 x}^{- 8} 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)]_{7 x}^{- 8}

Compute the boundaries:

\frac{π}{2} § (- 8 \frac{2}{π}) - \frac{π}{2} § (\frac{2}{π} \cdot 7 x)

= \frac{π}{2} § (- 8 \frac{2}{π}) - \frac{π}{2} § (\frac{2}{π} \cdot 7 x)

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What is the integral from 7x to-8 of sin(t^2) ?
The integral from 7x to-8 of sin(t^2) is sqrt(pi/2)\S(-8sqrt(2/pi))-sqrt(pi/2)\S(sqrt(2/pi)*7x)