What is the integral from 0 to pi/2 of cos^{2022}(x)sin(x) ?

The integral from 0 to pi/2 of cos^{2022}(x)sin(x) is 1/2023

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integral from 0 to pi/2 of cos^{2022}(x)sin(x)

\int_{0}^{\frac{π}{2}} cos^{2022} (x) sin (x) d x

\frac{1}{2023}

Decimal Notation

0.00049 \dots

Solution steps

\int_{0}^{\frac{π}{2}} cos^{2022} (x) sin (x) d x

Apply u-substitution

: \int_{1}^{0} - u^{2022} d u

= \int_{1}^{0} - u^{2022} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - u^{2022} d u

Take the constant out:

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

= - (- \int_{0}^{1} u^{2022} d u)

Apply the Power Rule

: [\frac{u ^{2023}}{2023}]_{0}^{1}

= - (- [\frac{u ^{2023}}{2023}]_{0}^{1})

Simplify

= [\frac{u ^{2023}}{2023}]_{0}^{1}

Compute the boundaries:

\frac{1}{2023}

= \frac{1}{2023}

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What is the integral from 0 to pi/2 of cos^{2022}(x)sin(x) ?
The integral from 0 to pi/2 of cos^{2022}(x)sin(x) is 1/2023