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

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

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

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

\frac{8}{5}

Solution steps

\int_{0}^{π} 4 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

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

Apply u-substitution

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

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

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

= 4 (- \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

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

Apply the Power Rule

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

= 4 (- (- [\frac{u ^{5}}{5}]_{- 1}^{1}))

Simplify

= 4 [\frac{u ^{5}}{5}]_{- 1}^{1}

Compute the boundaries:

\frac{2}{5}

= 4 \cdot \frac{2}{5}

Simplify

= \frac{8}{5}

18 e^{x} + 15 x

\frac{d}{d x} (3 e^{x + 1})

(ln (x^{2}))^{^{'}}

\frac{d}{d x} (\frac{( 2 )}{( 6 - 12 x )})

5 cos (π x), 12 x^{2} - 3, - 0.5, 0.5

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