What is the integral from 0 to 3pi of sin^5(x) ?

The integral from 0 to 3pi of sin^5(x) is 16/15

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integral from 0 to 3pi of sin^5(x)

\int_{0}^{3 π} sin^{5} (x) d x

\frac{16}{15}

Solution steps

\int_{0}^{3 π} sin^{5} (x) d x

Rewrite using trig identities

= \int_{0}^{3 π} (1 - cos^{2} (x))^{2} sin (x) d x

Apply u-substitution

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

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

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

= - \int_{- 1}^{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

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

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

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

\int_{- 1}^{1} u^{4} d u = \frac{2}{5}

= - (- 2 + \frac{4}{3} - \frac{2}{5})

Simplify

- (- 2 + \frac{4}{3} - \frac{2}{5}) : \frac{16}{15}

= \frac{16}{15}

\int_{0}^{\frac{π}{6}} x sin (6 x) d x

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

\int_{- 1}^{1} \frac{x}{cos ^{2} ( x )} d x

\int_{0}^{3} (4 x^{2} - 2 x + 5) d x

What is the integral from 0 to 3pi of sin^5(x) ?
The integral from 0 to 3pi of sin^5(x) is 16/15