What is the integral from 0 to pi/3 of sin^4(3x) ?

The integral from 0 to pi/3 of sin^4(3x) is pi/8

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integral from 0 to pi/3 of sin^4(3x)

\int_{0}^{\frac{π}{3}} sin^{4} (3 x) d x

\frac{π}{8}

Decimal

0.39269 \dots

Solution steps

\int_{0}^{\frac{π}{3}} sin^{4} (3 x) d x

Apply u-substitution

: \int_{0}^{π} sin^{4} (u) \frac{1}{3} d u

= \int_{0}^{π} sin^{4} (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{π} sin^{4} (u) d u

Apply Integral Reduction

= \frac{1}{3} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{π} + \frac{3}{4} \cdot \int_{0}^{π} sin^{2} (u) d u)

\int_{0}^{π} sin^{2} (u) d u = \frac{π}{2}

= \frac{1}{3} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2})

Simplify

\frac{1}{3} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2}) : - \frac{1}{3} [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{π} + \frac{π}{8}

= - \frac{1}{3} [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{π} + \frac{π}{8}

Compute the boundaries:

0

= - \frac{1}{3} \cdot 0 + \frac{π}{8}

Simplify

= \frac{π}{8}

ln (\frac{x + 3}{x - 1})

121 y^{^{''}} - 16 y = 0

\frac{\partial}{\partial x} ((x + y)^{2})

n = 1 \sum \infty \frac{1}{n ^{2} + 3 n + 2}

\int \frac{( x + 1 )}{( x ^{2} + 2 x )} d x

What is the integral from 0 to pi/3 of sin^4(3x) ?
The integral from 0 to pi/3 of sin^4(3x) is pi/8