What is the integral from x^2 to 0 of xcos(t^2) ?

The integral from x^2 to 0 of xcos(t^2) is x(sqrt(pi/2)\C(sqrt(2/pi)*0)-sqrt(pi/2)\C(sqrt(2/pi)x^2))

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integral from x^2 to 0 of xcos(t^2)

\int_{x^{2}}^{0} x cos (t^{2}) d t

x (\frac{π}{2} C (\frac{2}{π} \cdot 0) - \frac{π}{2} C (\frac{2}{π} x^{2}))

Solution steps

\int_{x^{2}}^{0} x cos (t^{2}) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= x \cdot \int_{x^{2}}^{0} cos (t^{2}) d t

Use the nonelementary integral:

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

= x [\frac{π}{2} C (\frac{2}{π} t)]_{x^{2}}^{0}

Compute the boundaries:

\frac{π}{2} C (\frac{2}{π} \cdot 0) - \frac{π}{2} C (\frac{2}{π} x^{2})

= x (\frac{π}{2} C (\frac{2}{π} \cdot 0) - \frac{π}{2} C (\frac{2}{π} x^{2}))

\int_{0}^{9} 9 - x d x

\int_{0}^{1} \frac{x}{x ^{2} - 9} d x

\int_{0}^{2} 6 e^{2 x} d x

\int_{- 4}^{4} (x^{2} + x - 6) d x

\int_{4}^{9} x x - 3 d x

What is the integral from x^2 to 0 of xcos(t^2) ?
The integral from x^2 to 0 of xcos(t^2) is x(sqrt(pi/2)\C(sqrt(2/pi)*0)-sqrt(pi/2)\C(sqrt(2/pi)x^2))