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}

Decimal Notation

3.2

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}

tan^{2} (3 x) + 1

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

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

\frac{d}{d x} (1.1 x + 6.3 x^{3})

\frac{d y}{d x} + 3 x y^{2} = 0

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