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

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

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

\int_{0}^{π} cos^{2} (x) sin (x) d x

\frac{2}{3}

Decimal

0.66666 \dots

Solution steps

\int_{0}^{π} cos^{2} (x) sin (x) d x

Apply u-substitution

: \int_{1}^{- 1} - u^{2} d u

= \int_{1}^{- 1} - u^{2} d u

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

= - \int_{- 1}^{1} - u^{2} d u

Take the constant out:

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

= - (- \int_{- 1}^{1} u^{2} d u)

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{- 1}^{1}

= - (- [\frac{u ^{3}}{3}]_{- 1}^{1})

Simplify

= [\frac{u ^{3}}{3}]_{- 1}^{1}

Compute the boundaries:

\frac{2}{3}

= \frac{2}{3}

2 cos^{3} (x)

\int (8 x - 2 x) d x

\int \frac{1}{1 + 2 sin ( x )} d x

\frac{d y}{d x} = y + 6

\int 12 x 1 + 6 x^{2} d x

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