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

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

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

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

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

\int_{- π}^{0} - 2 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

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

Apply u-substitution

= - 2 \cdot \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

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

Apply the Power Rule

= - 2 (- [\frac{u ^{3}}{3}]_{- 1}^{1})

Simplify

= 2 [\frac{u ^{3}}{3}]_{- 1}^{1}

Compute the boundaries:

\frac{2}{3}

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

Simplify

= \frac{4}{3}

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