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

The integral from 0 to pi of xsin(x)cos(x) is -pi/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to pi of xsin(x)cos(x)

\int_{0}^{π} x sin (x) cos (x) d x

- \frac{π}{4}

Decimal

- 0.78539 \dots

Solution steps

\int_{0}^{π} x sin (x) cos (x) d x

Rewrite using trig identities

= \int_{0}^{π} \frac{1}{2} x sin (2 x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{π} x sin (2 x) d x

Apply Integration By Parts

: [\frac{1}{2} (- x cos (2 x) - 2 \cdot \int - \frac{1}{2} cos (2 x) d x)]_{0}^{π}

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

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

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

Simplify

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

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

Compute the boundaries:

- \frac{π}{2}

= \frac{1}{2} (- \frac{π}{2})

Simplify

= - \frac{π}{4}

(3 x y + y^{2}) + (x^{2} + x y) \frac{d y}{d x} = 0

\int (tan^{2} (x)) d x

\int \frac{x}{x ^{2} + 6 x + 12} d x

4 t^{2} + 1

(1 + x^{2}) + (1 + y) y^{^{'}} = 0

What is the integral from 0 to pi of xsin(x)cos(x) ?
The integral from 0 to pi of xsin(x)cos(x) is -pi/4