What is the integral from 0 to 1 of 6cos(x^2) ?

The integral from 0 to 1 of 6cos(x^2) is 6(sqrt(pi/2)\C(sqrt(2/pi)*1)-sqrt(pi/2)\C(sqrt(2/pi)*0))

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integral from 0 to 1 of 6cos(x^2)

\int_{0}^{1} 6 cos (x^{2}) d x

6 (\frac{π}{2} C (\frac{2}{π} \cdot 1) - \frac{π}{2} C (\frac{2}{π} \cdot 0))

Solution steps

\int_{0}^{1} 6 cos (x^{2}) d x

Take the constant out:

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

= 6 \cdot \int_{0}^{1} cos (x^{2}) d x

Use the nonelementary integral:

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

= 6 [\frac{π}{2} C (\frac{2}{π} x)]_{0}^{1}

Compute the boundaries:

\frac{π}{2} C (\frac{2}{π} \cdot 1) - \frac{π}{2} C (\frac{2}{π} \cdot 0)

= 6 (\frac{π}{2} C (\frac{2}{π} \cdot 1) - \frac{π}{2} C (\frac{2}{π} \cdot 0))

\frac{\partial}{\partial x} (4 x^{3} - 4 y)

\frac{d y}{d x} + \frac{3 y}{x} = 6 x^{2}

sin (2 t)

\frac{\partial}{\partial x} (x^{0.75})

y^{^{''}} + 2 y^{^{'}} - y = 0, y (0) = 0, y^{^{'}} (0) = 22

What is the integral from 0 to 1 of 6cos(x^2) ?
The integral from 0 to 1 of 6cos(x^2) is 6(sqrt(pi/2)\C(sqrt(2/pi)*1)-sqrt(pi/2)\C(sqrt(2/pi)*0))