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

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

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

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

- 2

Solution steps

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

Take the constant out:

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

= - 3 \cdot \int_{0}^{π} cos^{2} (x) sin (x) d x

Apply u-substitution

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

= - 3 \cdot \int_{1}^{- 1} - u^{2} d u

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

= - 3 (- \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

= - 3 (- (- \int_{- 1}^{1} u^{2} d u))

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

\frac{2}{3}

= - 3 \cdot \frac{2}{3}

Simplify

= - 2

\int_{0}^{1} 8 cos (\frac{π t}{2}) d t

\int_{0}^{1} 6 x^{2} d x

\int_{0}^{π} x cos (3 x) d x

\int_{1}^{2} \frac{1}{1 - x} d x

\int_{0}^{π} 1 - sin^{3} (x) d x

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