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

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

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

\int_{0}^{\frac{π}{2}} sin^{3} (2 x) d x

\frac{2}{3}

Decimal

0.66666 \dots

Solution steps

\int_{0}^{\frac{π}{2}} sin^{3} (2 x) d x

Rewrite using trig identities

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

Rewrite using trig identities

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

Apply u-substitution

: \int_{1}^{0} - u^{3} (1 - u^{2}) d u

= 8 \cdot \int_{1}^{0} - u^{3} (1 - u^{2}) d u

Expand

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

= 8 \cdot \int_{1}^{0} - u^{3} + u^{5} d u

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

= 8 (- \int_{0}^{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_{0}^{1} u^{3} d u + \int_{0}^{1} u^{5} d u))

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

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

= 8 (- (- \frac{1}{4} + \frac{1}{6}))

Simplify

8 (- (- \frac{1}{4} + \frac{1}{6})) : \frac{2}{3}

= \frac{2}{3}

\frac{d}{d x} (e^{0.4 x} \cdot cos (0.2 x))

x^{2}, 8 - x^{2}

\int_{0}^{1} x f (x) d x

\frac{\partial}{\partial x} (2 x^{4} y - 3 x y + 9 y)

\frac{d y}{d x} = 3 x + y - 3

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