What is the integral from 0 to pi/8 of sin^5(2x)cos(2x) ?

The integral from 0 to pi/8 of sin^5(2x)cos(2x) is 1/96

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integral from 0 to pi/8 of sin^5(2x)cos(2x)

\int_{0}^{\frac{π}{8}} sin^{5} (2 x) cos (2 x) d x

\frac{1}{96}

Solution steps

\int_{0}^{\frac{π}{8}} sin^{5} (2 x) cos (2 x) d x

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: [\frac{5}{6} u^{\frac{6}{5}}]_{0}^{\frac{1}{4 2}}

= \frac{1}{10} [\frac{5}{6} u^{\frac{6}{5}}]_{0}^{\frac{1}{4 2}}

Compute the boundaries:

\frac{5}{48}

= \frac{1}{10} \cdot \frac{5}{48}

Simplify

= \frac{1}{96}

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What is the integral from 0 to pi/8 of sin^5(2x)cos(2x) ?
The integral from 0 to pi/8 of sin^5(2x)cos(2x) is 1/96