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

The integral from 0 to pi/2 of xcos^2(x) is (pi^2-2)/(16)-1/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{π ^{2} - 2}{16} - \frac{1}{8}

Decimal

0.36685 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 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}^{\frac{π}{2}}

\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}^{\frac{π}{2}}

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

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

Compute the boundaries:

\frac{π ^{2} - 2}{16} - \frac{1}{8}

= \frac{π ^{2} - 2}{16} - \frac{1}{8}

\int_{0}^{3} x^{2} - 1 d x

\int_{1}^{7} \frac{1}{x ^{2}} d x

\int_{1}^{4} u d u

\int_{0}^{1} x^{4} (1 + 2 x^{5})^{5} d x

\int_{- 1}^{0} - x d x

What is the integral from 0 to pi/2 of xcos^2(x) ?
The integral from 0 to pi/2 of xcos^2(x) is (pi^2-2)/(16)-1/8