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

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

Popular >

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

\int_{0}^{π} cos^{2} (t) sin (t) d t

\frac{2}{3}

Solution steps

\int_{0}^{π} cos^{2} (t) sin (t) d t

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}

\int_{0}^{\frac{π}{2}} y^{2} sin (y) d y

\int_{0}^{x^{2}} ln (1 + t) d t

\int_{- 2}^{0} \frac{1}{e ^{- 2 x}} d x

\int_{0}^{1} e^{x} + 1 d x

\int_{- 2}^{1} (x^{2} + x - 2) d x

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