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

The integral from 0 to 2pi of x(cos(x))^2 is pi^2

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integral from 0 to 2pi of x(cos(x))^2

\int_{0}^{2 π} x (cos (x))^{2} d x

π^{2}

Decimal

9.86960 \dots

Solution steps

\int_{0}^{2 π} x (cos (x))^{2} d x

Simplify

= \int_{0}^{2 π} x cos^{2} (x) d x

Apply Integration By Parts

= [\frac{1}{4} (x (2 x + sin (2 x)) - 4 \cdot \int \frac{1}{2} (x + \frac{1}{2} sin (2 x)) d x)]_{0}^{2 π}

\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}{4} (x (2 x + sin (2 x)) - 4 \cdot \frac{1}{2} (\frac{x ^{2}}{2} - \frac{1}{4} cos (2 x)))]_{0}^{2 π}

Simplify

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

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

Compute the boundaries:

\frac{8 π ^{2} + 1}{8} - \frac{1}{8}

= \frac{8 π ^{2} + 1}{8} - \frac{1}{8}

Simplify

= π^{2}

What is the integral from 0 to 2pi of x(cos(x))^2 ?
The integral from 0 to 2pi of x(cos(x))^2 is pi^2