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

The integral from 0 to pi of x^2sin(kx) is (-pi^2k^2cos(pik)+2(piksin(pik)+cos(pik))-2)/(k^3)

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(kx)

\int_{0}^{π} x^{2} sin (k x) d x

\frac{- π ^{2} k ^{2} cos ( πk ) + 2 ( πk sin ( πk ) + cos ( πk ) ) - 2}{k ^{3}}

Solution steps

\int_{0}^{π} x^{2} sin (k x) d x

Apply Integration By Parts

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

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

\int - \frac{2}{k} x cos (k x) d x = - \frac{2}{k ^{3}} (k x sin (k x) + cos (k x))

= [\frac{1}{k} (- x^{2} cos (k x) - k (- \frac{2}{k ^{3}} (k x sin (k x) + cos (k x))))]_{0}^{π}

Simplify

[\frac{1}{k} (- x^{2} cos (k x) - k (- \frac{2}{k ^{3}} (k x sin (k x) + cos (k x))))]_{0}^{π} : [\frac{1}{k} (- x^{2} cos (k x) + \frac{2}{k ^{2}} (k x sin (k x) + cos (k x)))]_{0}^{π}

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

Compute the boundaries:

\frac{1}{k} (- π^{2} cos (kπ) + \frac{2}{k ^{2}} (kπ sin (kπ) + cos (kπ))) - \frac{2}{k ^{3}}

= \frac{1}{k} (- π^{2} cos (kπ) + \frac{2}{k ^{2}} (kπ sin (kπ) + cos (kπ))) - \frac{2}{k ^{3}}

Simplify

= \frac{- π ^{2} k ^{2} cos ( πk ) + 2 ( πk sin ( πk ) + cos ( πk ) ) - 2}{k ^{3}}

\int_{- 2}^{6} (3 - \frac{x ^{2}}{4} + x) d x

\int_{1}^{4} (3 x^{2} - 2 x + 4) d x

\int_{1}^{2} e^{x} + 2 d x

\int_{1}^{2} \frac{3 x - y}{9} d y

\int_{0}^{0.5} \frac{1}{1 - x} d x

What is the integral from 0 to pi of x^2sin(kx) ?
The integral from 0 to pi of x^2sin(kx) is (-pi^2k^2cos(pik)+2(piksin(pik)+cos(pik))-2)/(k^3)