What is the integral from 3x to-5 of sin(t^2) ?

The integral from 3x to-5 of sin(t^2) is sqrt(pi/2)\S(-5sqrt(2/pi))-sqrt(pi/2)\S(sqrt(2/pi)*3x)

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integral from 3x to-5 of sin(t^2)

\int_{3 x}^{- 5} sin (t^{2}) d t

\frac{π}{2} § (- 5 \frac{2}{π}) - \frac{π}{2} § (\frac{2}{π} \cdot 3 x)

Solution steps

\int_{3 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)]_{3 x}^{- 5}

Compute the boundaries:

\frac{π}{2} § (- 5 \frac{2}{π}) - \frac{π}{2} § (\frac{2}{π} \cdot 3 x)

= \frac{π}{2} § (- 5 \frac{2}{π}) - \frac{π}{2} § (\frac{2}{π} \cdot 3 x)

\int_{0}^{2} e^{x^{2} - 2 x} d x

\int_{0}^{1} (x)^{2} d x

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

\int_{2}^{5} 6 x d x

\int_{1}^{5} ∣ 2 x - 6 ∣ d x

What is the integral from 3x to-5 of sin(t^2) ?
The integral from 3x to-5 of sin(t^2) is sqrt(pi/2)\S(-5sqrt(2/pi))-sqrt(pi/2)\S(sqrt(2/pi)*3x)