What is the integral from 0 to pi of 8cos^4(x)sin(x) ?

The integral from 0 to pi of 8cos^4(x)sin(x) is 16/5

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integral from 0 to pi of 8cos^4(x)sin(x)

\int_{0}^{π} 8 cos^{4} (x) sin (x) d x

\frac{16}{5}

Decimal Notation

3.2

Solution steps

\int_{0}^{π} 8 cos^{4} (x) sin (x) d x

Take the constant out:

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

= 8 \cdot \int_{0}^{π} cos^{4} (x) sin (x) d x

Apply u-substitution

: \int_{1}^{- 1} - u^{4} d u

= 8 \cdot \int_{1}^{- 1} - u^{4} d u

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

= 8 (- \int_{- 1}^{1} - u^{4} d u)

Take the constant out:

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

= 8 (- (- \int_{- 1}^{1} u^{4} d u))

Apply the Power Rule

: [\frac{u ^{5}}{5}]_{- 1}^{1}

= 8 (- (- [\frac{u ^{5}}{5}]_{- 1}^{1}))

Simplify

= 8 [\frac{u ^{5}}{5}]_{- 1}^{1}

Compute the boundaries:

\frac{2}{5}

= 8 \cdot \frac{2}{5}

Simplify

= \frac{16}{5}

\int 3 x^{2} + 1 d x

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

\frac{d y}{d x} = \frac{x ^{2} y - y}{y + 1}

\int \frac{1}{18} x d x

What is the integral from 0 to pi of 8cos^4(x)sin(x) ?
The integral from 0 to pi of 8cos^4(x)sin(x) is 16/5