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

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

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

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

\frac{128}{315}

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \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}

\int_{- 2}^{x} (20 t^{2} - 19 t) d t

\int_{\frac{π}{2}}^{\frac{3 π}{2}} 2 cos (x) d x

\int_{0}^{2} \frac{1}{1 + x ^{4}} d x

\int_{0}^{1} - t d t

\int_{- 1}^{x} e^{y^{2}} d y

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