What is the integral from 0 to pi/8 of sin^3(8x) ?

The integral from 0 to pi/8 of sin^3(8x) is 1/6

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integral from 0 to pi/8 of sin^3(8x)

\int_{0}^{\frac{π}{8}} sin^{3} (8 x) d x

\frac{1}{6}

Decimal

0.16666 \dots

Solution steps

\int_{0}^{\frac{π}{8}} sin^{3} (8 x) d x

Apply u-substitution

: \int_{0}^{π} sin^{3} (u) \frac{1}{8} d u

= \int_{0}^{π} sin^{3} (u) \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{0}^{π} sin^{3} (u) d u

Rewrite using trig identities

= \frac{1}{8} \cdot \int_{0}^{π} (1 - cos^{2} (u)) sin (u) d u

Apply u-substitution

: \int_{1}^{- 1} - 1 + v^{2} d v

= \frac{1}{8} \cdot \int_{1}^{- 1} - 1 + v^{2} d v

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

= \frac{1}{8} (- \int_{- 1}^{1} - 1 + v^{2} d v)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (- (- \int_{- 1}^{1} 1 d v + \int_{- 1}^{1} v^{2} d v))

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

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

= \frac{1}{8} (- (- 2 + \frac{2}{3}))

Simplify

\frac{1}{8} (- (- 2 + \frac{2}{3})) : \frac{1}{6}

= \frac{1}{6}

\int_{- \infty}^{2} x^{2} d x

\int_{- 4}^{3} x d x

\int_{1}^{3} x^{2} - 8 x + 15 d x

\int_{1}^{4} \frac{ln ( x )}{x ^{4}} d x

\int_{0}^{4 - y^{2}} 1

What is the integral from 0 to pi/8 of sin^3(8x) ?
The integral from 0 to pi/8 of sin^3(8x) is 1/6