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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

\int_{0}^{2 π} x sin^{2} (x) d x

π^{2}

Decimal

9.86960 \dots

Solution steps

\int_{0}^{2 π} x sin^{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 xsin^2(x) ?
The integral from 0 to 2pi of xsin^2(x) is pi^2