What is the integral from 0 to 4 of xcos^2(x) ?

The integral from 0 to 4 of xcos^2(x) is sin(8)+31/8+1/8 cos(8)

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integral from 0 to 4 of xcos^2(x)

\int_{0}^{4} x cos^{2} (x) d x

sin (8) + \frac{31}{8} + \frac{1}{8} cos (8)

Decimal

4.84617 \dots

Solution steps

\int_{0}^{4} 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]_{0}^{4}

\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))]_{0}^{4}

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))]_{0}^{4} : [\frac{1}{8} (2 x^{2} + 2 x sin (2 x) + cos (2 x))]_{0}^{4}

= [\frac{1}{8} (2 x^{2} + 2 x sin (2 x) + cos (2 x))]_{0}^{4}

Compute the boundaries:

sin (8) + \frac{31}{8} + \frac{1}{8} cos (8)

= sin (8) + \frac{31}{8} + \frac{1}{8} cos (8)

What is the integral from 0 to 4 of xcos^2(x) ?
The integral from 0 to 4 of xcos^2(x) is sin(8)+31/8+1/8 cos(8)