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

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

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

\int_{0}^{π} sin (t) cos^{2} (t) d t

\frac{2}{3}

Decimal

0.66666 \dots

Solution steps

\int_{0}^{π} sin (t) cos^{2} (t) d t

Apply u-substitution

= \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})

Simplify

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

Compute the boundaries:

\frac{2}{3}

= \frac{2}{3}

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