What is the integral from 0 to 2 of x^2cos((npix)/2) ?

The integral from 0 to 2 of x^2cos((npix)/2) is (16(-1)^n)/(pi^2n^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of x^2cos((npix)/2)

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

\frac{16 ( - 1 ) ^{n}}{π ^{2} n ^{2}}

Solution steps

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

Apply Integration By Parts

: [\frac{2}{πn} x^{2} sin (\frac{πn x}{2}) - \int \frac{4}{πn} x sin (\frac{πn x}{2}) d x]_{0}^{2}

= [\frac{2}{πn} x^{2} sin (\frac{πn x}{2}) - \int \frac{4}{πn} x sin (\frac{πn x}{2}) d x]_{0}^{2}

\int \frac{4}{πn} x sin (\frac{πn x}{2}) d x = \frac{16}{π ^{3} n ^{3}} (- \frac{πn}{2} x cos (\frac{πn x}{2}) + sin (\frac{πn x}{2}))

= [\frac{2}{πn} x^{2} sin (\frac{πn x}{2}) - \frac{16}{π ^{3} n ^{3}} (- \frac{πn}{2} x cos (\frac{πn x}{2}) + sin (\frac{πn x}{2}))]_{0}^{2}

Compute the boundaries:

\frac{16 ( - 1 ) ^{n}}{π ^{2} n ^{2}}

= \frac{16 ( - 1 ) ^{n}}{π ^{2} n ^{2}}

\int_{0}^{e + 1} \frac{x + 1}{x ^{2} - 1} d x

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

\int_{0}^{π} sin (x + n x) d x

\int_{0}^{2} x + 2 - x d x

\int_{2}^{4} \frac{x - 2}{x ^{2} - 5 x + 4} d x

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