What is the integral from 0 to pi/2 of x^3cos(x^2) ?

The integral from 0 to pi/2 of x^3cos(x^2) is 1/2 ((pi^2)/4 sin((pi^2)/4)+cos((pi^2)/4)-1)

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integral from 0 to pi/2 of x^3cos(x^2)

\int_{0}^{\frac{π}{2}} x^{3} cos (x^{2}) d x

\frac{1}{2} (\frac{π ^{2}}{4} sin (\frac{π ^{2}}{4}) + cos (\frac{π ^{2}}{4}) - 1)

Decimal

- 0.12044 \dots

Solution steps

\int_{0}^{\frac{π}{2}} x^{3} cos (x^{2}) d x

Apply u-substitution

: \int_{0}^{\frac{π ^{2}}{4}} \frac{u cos ( u )}{2} d u

= \int_{0}^{\frac{π ^{2}}{4}} \frac{u cos ( u )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{π ^{2}}{4}} u cos (u) d u

Apply Integration By Parts

: [u sin (u) - \int sin (u) d u]_{0}^{\frac{π ^{2}}{4}}

= \frac{1}{2} [u sin (u) - \int sin (u) d u]_{0}^{\frac{π ^{2}}{4}}

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

= \frac{1}{2} [u sin (u) - (- cos (u))]_{0}^{\frac{π ^{2}}{4}}

Simplify

= \frac{1}{2} [u sin (u) + cos (u)]_{0}^{\frac{π ^{2}}{4}}

Compute the boundaries:

\frac{π ^{2}}{4} sin (\frac{π ^{2}}{4}) + cos (\frac{π ^{2}}{4}) - 1

= \frac{1}{2} (\frac{π ^{2}}{4} sin (\frac{π ^{2}}{4}) + cos (\frac{π ^{2}}{4}) - 1)

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What is the integral from 0 to pi/2 of x^3cos(x^2) ?
The integral from 0 to pi/2 of x^3cos(x^2) is 1/2 ((pi^2)/4 sin((pi^2)/4)+cos((pi^2)/4)-1)