What is the integral from 0 to pi of sin^3(2x) ?

The integral from 0 to pi of sin^3(2x) is 0

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integral from 0 to pi of sin^3(2x)

\int_{0}^{π} sin^{3} (2 x) d x

0

Solution steps

\int_{0}^{π} sin^{3} (2 x) d x

Rewrite using trig identities

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

Take the constant out:

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

= 8 \cdot \int_{0}^{π} cos^{3} (x) sin^{3} (x) d x

Rewrite using trig identities

= 8 \cdot \int_{0}^{π} (1 - cos^{2} (x)) sin (x) cos^{3} (x) d x

Apply u-substitution

: \int_{1}^{- 1} - u^{3} (1 - u^{2}) d u

= 8 \cdot \int_{1}^{- 1} - u^{3} (1 - u^{2}) d u

Expand

- u^{3} (1 - u^{2}) : - u^{3} + u^{5}

= 8 \cdot \int_{1}^{- 1} - u^{3} + u^{5} d u

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

= 8 (- \int_{- 1}^{1} - u^{3} + u^{5} 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

= 8 (- (- \int_{- 1}^{1} u^{3} d u + \int_{- 1}^{1} u^{5} d u))

\int_{- 1}^{1} u^{3} d u = 0

\int_{- 1}^{1} u^{5} d u = 0

= 8 (- (- 0 + 0))

Simplify

= 0

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What is the integral from 0 to pi of sin^3(2x) ?
The integral from 0 to pi of sin^3(2x) is 0