What is the integral from 0 to pi of 6cos^4(x)sin(x) ?

The integral from 0 to pi of 6cos^4(x)sin(x) is 12/5

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integral from 0 to pi of 6cos^4(x)sin(x)

\int_{0}^{π} 6 cos^{4} (x) sin (x) d x

\frac{12}{5}

Decimal Notation

2.4

Solution steps

\int_{0}^{π} 6 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

= 6 \cdot \int_{0}^{π} cos^{4} (x) sin (x) d x

Apply u-substitution

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

= 6 \cdot \int_{1}^{- 1} - u^{4} d u

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

= 6 (- \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

= 6 (- (- \int_{- 1}^{1} u^{4} d u))

Apply the Power Rule

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

= 6 (- (- [\frac{u ^{5}}{5}]_{- 1}^{1}))

Simplify

= 6 [\frac{u ^{5}}{5}]_{- 1}^{1}

Compute the boundaries:

\frac{2}{5}

= 6 \cdot \frac{2}{5}

Simplify

= \frac{12}{5}

x \to - 0.1 lim (1 + e^{x})

4 y - y^{^{''}} = 10 e^{(- 3 x)}

e^{2 x + y} y^{^{'}} = e^{x - y}

\frac{d}{d x} (x^{5} - 5 x^{3})

\frac{e ^{x}}{x ^{2} + 3}

What is the integral from 0 to pi of 6cos^4(x)sin(x) ?
The integral from 0 to pi of 6cos^4(x)sin(x) is 12/5