What is the integral from 0 to pi/4 of 5xcos(x^2) ?

The integral from 0 to pi/4 of 5xcos(x^2) is 5/2 sin((pi^2)/(16))

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integral from 0 to pi/4 of 5xcos(x^2)

\int_{0}^{\frac{π}{4}} 5 x cos (x^{2}) d x

\frac{5}{2} sin (\frac{π ^{2}}{16})

Decimal

1.44617 \dots

Solution steps

\int_{0}^{\frac{π}{4}} 5 x cos (x^{2}) d x

Take the constant out:

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

= 5 \cdot \int_{0}^{\frac{π}{4}} x cos (x^{2}) d x

Apply u-substitution

: \int_{0}^{\frac{π ^{2}}{16}} \frac{cos ( u )}{2} d u

= 5 \cdot \int_{0}^{\frac{π ^{2}}{16}} \frac{cos ( u )}{2} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π ^{2}}{16}} cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= 5 \cdot \frac{1}{2} [sin (u)]_{0}^{\frac{π ^{2}}{16}}

Simplify

5 \cdot \frac{1}{2} [sin (u)]_{0}^{\frac{π ^{2}}{16}} : \frac{5}{2} [sin (u)]_{0}^{\frac{π ^{2}}{16}}

= \frac{5}{2} [sin (u)]_{0}^{\frac{π ^{2}}{16}}

Compute the boundaries:

sin (\frac{π ^{2}}{16})

= \frac{5}{2} sin (\frac{π ^{2}}{16})

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What is the integral from 0 to pi/4 of 5xcos(x^2) ?
The integral from 0 to pi/4 of 5xcos(x^2) is 5/2 sin((pi^2)/(16))