What is the integral from 0 to pi/2 of cos^9(x) ?

The integral from 0 to pi/2 of cos^9(x) is 128/315

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integral from 0 to pi/2 of cos^9(x)

\int_{0}^{\frac{π}{2}} cos^{9} (x) d x

\frac{128}{315}

Decimal Notation

0.40634 \dots

Solution steps

\int_{0}^{\frac{π}{2}} cos^{9} (x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{2}} (1 - sin^{2} (x))^{4} cos (x) d x

Apply u-substitution

: \int_{0}^{1} (1 - u^{2})^{4} d u

= \int_{0}^{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_{0}^{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_{0}^{1} 1 d u - \int_{0}^{1} 4 u^{2} d u + \int_{0}^{1} 6 u^{4} d u - \int_{0}^{1} 4 u^{6} d u + \int_{0}^{1} u^{8} d u

\int_{0}^{1} 1 d u = 1

\int_{0}^{1} 4 u^{2} d u = \frac{4}{3}

\int_{0}^{1} 6 u^{4} d u = \frac{6}{5}

\int_{0}^{1} 4 u^{6} d u = \frac{4}{7}

\int_{0}^{1} u^{8} d u = \frac{1}{9}

= 1 - \frac{4}{3} + \frac{6}{5} - \frac{4}{7} + \frac{1}{9}

Simplify

1 - \frac{4}{3} + \frac{6}{5} - \frac{4}{7} + \frac{1}{9} : \frac{128}{315}

= \frac{128}{315}

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

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

x \to - 1 lim (2 x^{2} + 3 x + 1)

\int 21 y d y

y^{^{''}} - \frac{2}{r} y^{^{'}} + \frac{2}{r ^{2}} y = 0, y (1) = 1, y (2) = 0

What is the integral from 0 to pi/2 of cos^9(x) ?
The integral from 0 to pi/2 of cos^9(x) is 128/315