What is the integral from 0 to pi/2 of 17cos^5(x) ?

The integral from 0 to pi/2 of 17cos^5(x) is 136/15

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integral from 0 to pi/2 of 17cos^5(x)

\int_{0}^{\frac{π}{2}} 17 cos^{5} (x) d x

\frac{136}{15}

Decimal Notation

9.06666 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 17 cos^{5} (x) d x

Take the constant out:

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

= 17 \cdot \int_{0}^{\frac{π}{2}} cos^{5} (x) d x

Rewrite using trig identities

= 17 \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

= 17 \cdot \int_{0}^{1} (1 - u^{2})^{2} d u

Expand

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

= 17 \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

= 17 (\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}

= 17 (1 - \frac{2}{3} + \frac{1}{5})

Simplify

17 (1 - \frac{2}{3} + \frac{1}{5}) : \frac{136}{15}

= \frac{136}{15}

\int \frac{2 x}{x ^{2} - 3 x - 10} d x

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

\frac{d}{d t} (\frac{cos ( t )}{sin ( t )})

\frac{d}{d x} (x arctan (2 x))

x \to 0 - lim (x csc (x))

What is the integral from 0 to pi/2 of 17cos^5(x) ?
The integral from 0 to pi/2 of 17cos^5(x) is 136/15