What is the integral from 0 to pi/2 of 16cos^5(x) ?

The integral from 0 to pi/2 of 16cos^5(x) is 128/15

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integral from 0 to pi/2 of 16cos^5(x)

\int_{0}^{\frac{π}{2}} 16 cos^{5} (x) d x

\frac{128}{15}

Decimal Notation

8.53333 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 16 cos^{5} (x) d x

Take the constant out:

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

= 16 \cdot \int_{0}^{\frac{π}{2}} cos^{5} (x) d x

Rewrite using trig identities

= 16 \cdot \int_{0}^{\frac{π}{2}} (1 - sin^{2} (x))^{2} cos (x) d x

Apply u-substitution

: \int_{0}^{1} (1 - u^{2})^{2} d u

= 16 \cdot \int_{0}^{1} (1 - u^{2})^{2} d u

Expand

(1 - u^{2})^{2} : 1 - 2 u^{2} + u^{4}

= 16 \cdot \int_{0}^{1} 1 - 2 u^{2} + u^{4} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 16 (\int_{0}^{1} 1 d u - \int_{0}^{1} 2 u^{2} d u + \int_{0}^{1} u^{4} d u)

\int_{0}^{1} 1 d u = 1

\int_{0}^{1} 2 u^{2} d u = \frac{2}{3}

\int_{0}^{1} u^{4} d u = \frac{1}{5}

= 16 (1 - \frac{2}{3} + \frac{1}{5})

Simplify

16 (1 - \frac{2}{3} + \frac{1}{5}) : \frac{128}{15}

= \frac{128}{15}

ln (x e^{x} + 4)

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

x \to \infty lim (x sin (\frac{3 π}{x}))

\int_{1}^{4} y ln (y) d y

\frac{d x}{d t} = k x (1000 - x)

What is the integral from 0 to pi/2 of 16cos^5(x) ?
The integral from 0 to pi/2 of 16cos^5(x) is 128/15