What is the integral from 13 to 16 of xcos^2(x) ?

The integral from 13 to 16 of xcos^2(x) is (174+32sin(32)+cos(32)-26sin(26)-cos(26))/8

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integral from 13 to 16 of xcos^2(x)

\int_{13}^{16} x cos^{2} (x) d x

\frac{174 + 32 sin ( 32 ) + cos ( 32 ) - 26 sin ( 26 ) - cos ( 26 )}{8}

Decimal

21.50080 \dots

Solution steps

\int_{13}^{16} x cos^{2} (x) d x

Apply Integration By Parts

: [\frac{1}{2} x (x + \frac{1}{2} sin (2 x)) - \int \frac{1}{2} (x + \frac{1}{2} sin (2 x)) d x]_{13}^{16}

= [\frac{1}{2} x (x + \frac{1}{2} sin (2 x)) - \int \frac{1}{2} (x + \frac{1}{2} sin (2 x)) d x]_{13}^{16}

\int \frac{1}{2} (x + \frac{1}{2} sin (2 x)) d x = \frac{1}{2} (\frac{x ^{2}}{2} - \frac{1}{4} cos (2 x))

= [\frac{1}{2} x (x + \frac{1}{2} sin (2 x)) - \frac{1}{2} (\frac{x ^{2}}{2} - \frac{1}{4} cos (2 x))]_{13}^{16}

Simplify

[\frac{1}{2} x (x + \frac{1}{2} sin (2 x)) - \frac{1}{2} (\frac{x ^{2}}{2} - \frac{1}{4} cos (2 x))]_{13}^{16} : [\frac{1}{8} (2 x^{2} + 2 x sin (2 x) + cos (2 x))]_{13}^{16}

= [\frac{1}{8} (2 x^{2} + 2 x sin (2 x) + cos (2 x))]_{13}^{16}

Compute the boundaries:

\frac{174 + 32 sin ( 32 ) + cos ( 32 ) - 26 sin ( 26 ) - cos ( 26 )}{8}

= \frac{174 + 32 sin ( 32 ) + cos ( 32 ) - 26 sin ( 26 ) - cos ( 26 )}{8}

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What is the integral from 13 to 16 of xcos^2(x) ?
The integral from 13 to 16 of xcos^2(x) is (174+32sin(32)+cos(32)-26sin(26)-cos(26))/8