What is the integral from-pi/2 to pi/2 of cos^9(x) ?

The integral from-pi/2 to pi/2 of cos^9(x) is 256/315

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integral from-pi/2 to pi/2 of cos^9(x)

\int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{9} (x) d x

\frac{256}{315}

Decimal

0.81269 \dots

Solution steps

\int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{9} (x) d x

Rewrite using trig identities

= \int_{- \frac{π}{2}}^{\frac{π}{2}} (1 - sin^{2} (x))^{4} cos (x) d x

Apply u-substitution

: \int_{- 1}^{1} (1 - u^{2})^{4} d u

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

Expand

(1 - u^{2})^{4} : 1 - 4 u^{2} + 6 u^{4} - 4 u^{6} + u^{8}

= \int_{- 1}^{1} 1 - 4 u^{2} + 6 u^{4} - 4 u^{6} + u^{8} 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

= \int_{- 1}^{1} 1 d u - \int_{- 1}^{1} 4 u^{2} d u + \int_{- 1}^{1} 6 u^{4} d u - \int_{- 1}^{1} 4 u^{6} d u + \int_{- 1}^{1} u^{8} d u

\int_{- 1}^{1} 1 d u = 2

\int_{- 1}^{1} 4 u^{2} d u = \frac{8}{3}

\int_{- 1}^{1} 6 u^{4} d u = \frac{12}{5}

\int_{- 1}^{1} 4 u^{6} d u = \frac{8}{7}

\int_{- 1}^{1} u^{8} d u = \frac{2}{9}

= 2 - \frac{8}{3} + \frac{12}{5} - \frac{8}{7} + \frac{2}{9}

Simplify

2 - \frac{8}{3} + \frac{12}{5} - \frac{8}{7} + \frac{2}{9} : \frac{256}{315}

= \frac{256}{315}

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What is the integral from-pi/2 to pi/2 of cos^9(x) ?
The integral from-pi/2 to pi/2 of cos^9(x) is 256/315