What is the integral from x to 6x of cos(t^2) ?

The integral from x to 6x of cos(t^2) is sqrt(pi/2)\C(sqrt(2/pi)*6x)-sqrt(pi/2)\C(sqrt(2/pi)x)

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integral from x to 6x of cos(t^2)

\int_{x}^{6 x} cos (t^{2}) d t

\frac{π}{2} C (\frac{2}{π} \cdot 6 x) - \frac{π}{2} C (\frac{2}{π} x)

Solution steps

\int_{x}^{6 x} cos (t^{2}) d t

Use the nonelementary integral:

\int cos (x^{2}) d x = \frac{π}{2} C (\frac{2}{π} x)

= [\frac{π}{2} C (\frac{2}{π} t)]_{x}^{6 x}

Compute the boundaries:

\frac{π}{2} C (\frac{2}{π} \cdot 6 x) - \frac{π}{2} C (\frac{2}{π} x)

= \frac{π}{2} C (\frac{2}{π} \cdot 6 x) - \frac{π}{2} C (\frac{2}{π} x)

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What is the integral from x to 6x of cos(t^2) ?
The integral from x to 6x of cos(t^2) is sqrt(pi/2)\C(sqrt(2/pi)*6x)-sqrt(pi/2)\C(sqrt(2/pi)x)