What is the integral from 0 to pi/2 of-6cos^5(x) ?

The integral from 0 to pi/2 of-6cos^5(x) is -16/5

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integral from 0 to pi/2 of-6cos^5(x)

\int_{0}^{\frac{π}{2}} - 6 cos^{5} (x) d x

- \frac{16}{5}

Solution steps

\int_{0}^{\frac{π}{2}} - 6 cos^{5} (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}^{\frac{π}{2}} cos^{5} (x) d x

Rewrite using trig identities

= - 6 \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

= - 6 \cdot \int_{0}^{1} (1 - u^{2})^{2} d u

Expand

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

= - 6 \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

= - 6 (\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}

= - 6 (1 - \frac{2}{3} + \frac{1}{5})

Simplify

- 6 (1 - \frac{2}{3} + \frac{1}{5}) : - \frac{16}{5}

= - \frac{16}{5}

\frac{d}{d x} (x^{2} ln (\frac{1}{x}))

\frac{\partial}{\partial x} (9 cos (x y) x)

x^{^{''}} + k^{2} x = 0

sin^{2} (x^{2} - 1)

\int (e^{\frac{1}{x}}) d x

What is the integral from 0 to pi/2 of-6cos^5(x) ?
The integral from 0 to pi/2 of-6cos^5(x) is -16/5