What is the integral from 0 to 5 of sin(t^2) ?

The integral from 0 to 5 of sin(t^2) is sqrt(pi/2)\S(sqrt(2/pi)*5)-sqrt(pi/2)\S(sqrt(2/pi)*0)

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integral from 0 to 5 of sin(t^2)

\int_{0}^{5} sin (t^{2}) d t

\frac{π}{2} § (\frac{2}{π} \cdot 5) - \frac{π}{2} § (\frac{2}{π} \cdot 0)

Solution steps

\int_{0}^{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)]_{0}^{5}

Compute the boundaries:

\frac{π}{2} § (\frac{2}{π} \cdot 5) - \frac{π}{2} § (\frac{2}{π} \cdot 0)

= \frac{π}{2} § (\frac{2}{π} \cdot 5) - \frac{π}{2} § (\frac{2}{π} \cdot 0)

\int_{3}^{6} 3 d x

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\int_{0}^{1} \frac{2 x + 1}{x ^{2} + 1} d x

\int_{2}^{5} \frac{k}{x ^{3}} d x

\int_{0}^{2} (x - x^{2}) d x

What is the integral from 0 to 5 of sin(t^2) ?
The integral from 0 to 5 of sin(t^2) is sqrt(pi/2)\S(sqrt(2/pi)*5)-sqrt(pi/2)\S(sqrt(2/pi)*0)