What is the integral from 0 to pi/(12) of xcos^2(6x) ?

The integral from 0 to pi/(12) of xcos^2(6x) is (pi^2-4)/(576)

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integral from 0 to pi/(12) of xcos^2(6x)

\int_{0}^{\frac{π}{12}} x cos^{2} (6 x) d x

\frac{π ^{2} - 4}{576}

Decimal

0.01019 \dots

Solution steps

\int_{0}^{\frac{π}{12}} x cos^{2} (6 x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{12}} \frac{1}{2} x (1 + cos (12 x)) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{12}} x (1 + cos (12 x)) d x

Apply Integration By Parts

: [x (x + \frac{1}{12} sin (12 x)) - \int \frac{1}{12} (12 x + sin (12 x)) d x]_{0}^{\frac{π}{12}}

= \frac{1}{2} [x (x + \frac{1}{12} sin (12 x)) - \int \frac{1}{12} (12 x + sin (12 x)) d x]_{0}^{\frac{π}{12}}

\int \frac{1}{12} (12 x + sin (12 x)) d x = \frac{1}{144} (72 x^{2} - cos (12 x))

= \frac{1}{2} [x (x + \frac{1}{12} sin (12 x)) - \frac{1}{144} (72 x^{2} - cos (12 x))]_{0}^{\frac{π}{12}}

Simplify

\frac{1}{2} [x (x + \frac{1}{12} sin (12 x)) - \frac{1}{144} (72 x^{2} - cos (12 x))]_{0}^{\frac{π}{12}} : \frac{1}{2} [\frac{72 x ^{2} + 12 x sin ( 12 x ) + cos ( 12 x )}{144}]_{0}^{\frac{π}{12}}

= \frac{1}{2} [\frac{72 x ^{2} + 12 x sin ( 12 x ) + cos ( 12 x )}{144}]_{0}^{\frac{π}{12}}

Compute the boundaries:

\frac{π ^{2} - 2}{288} - \frac{1}{144}

= \frac{1}{2} (\frac{π ^{2} - 2}{288} - \frac{1}{144})

Simplify

= \frac{π ^{2} - 4}{576}

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What is the integral from 0 to pi/(12) of xcos^2(6x) ?
The integral from 0 to pi/(12) of xcos^2(6x) is (pi^2-4)/(576)