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

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

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

- \frac{π ^{2}}{4}

Solution steps

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

Rewrite using trig identities

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{- 2 π ^{2} + 1}{4} - \frac{1}{4}

= \frac{1}{2} (\frac{- 2 π ^{2} + 1}{4} - \frac{1}{4})

Simplify

= - \frac{π ^{2}}{4}

\int_{- 2}^{2} x (x^{2} - 5) d x

\int_{- π}^{π} x^{3} d x

\int_{0}^{100} x d x

\int_{- 3}^{- 2} \frac{( x + 2 )}{x ^{2} + 4 x} d x

\int_{0}^{5} t 25 - t^{2} d t

What is the integral from 0 to pi of x^2sin(x)cos(x) ?
The integral from 0 to pi of x^2sin(x)cos(x) is -(pi^2)/4