What is the integral from 0 to pi/2 of 10cos^5(x) ?

The integral from 0 to pi/2 of 10cos^5(x) is 16/3

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integral from 0 to pi/2 of 10cos^5(x)

\int_{0}^{\frac{π}{2}} 10 cos^{5} (x) d x

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 10 cos^{5} (x) d x

Take the constant out:

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

= 10 \cdot \int_{0}^{\frac{π}{2}} cos^{5} (x) d x

Rewrite using trig identities

= 10 \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

= 10 \cdot \int_{0}^{1} (1 - u^{2})^{2} d u

Expand

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

= 10 \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

= 10 (\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}

= 10 (1 - \frac{2}{3} + \frac{1}{5})

Simplify

10 (1 - \frac{2}{3} + \frac{1}{5}) : \frac{16}{3}

= \frac{16}{3}

\int 3 x^{\frac{4}{3}} - sin (x) + c d x

\frac{d}{d x} (\frac{e ^{3 x}}{x ^{6}})

\int \frac{1}{( x + 2 ) ( x + 3 )} d x

\int \frac{5 x ^{2} + 3 x - 2}{x ^{3} + 2 x ^{2}} d x

x \to 0 lim (\frac{4 x}{x})

What is the integral from 0 to pi/2 of 10cos^5(x) ?
The integral from 0 to pi/2 of 10cos^5(x) is 16/3