What is the integral from 0 to 4 of xcos((npix)/4) ?

The integral from 0 to 4 of xcos((npix)/4) is (16(-1)^n-16)/(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 4 of xcos((npix)/4)

\int_{0}^{4} x cos (\frac{nπ x}{4}) d x

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

Solution steps

\int_{0}^{4} x cos (\frac{nπ x}{4}) d x

Apply u-substitution

: \int_{0}^{nπ 4} \frac{u cos ( \frac{u}{4} )}{π ^{2} n ^{2}} d u

= \int_{0}^{nπ 4} \frac{u cos ( \frac{u}{4} )}{π ^{2} n ^{2}} d u

Take the constant out:

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

= \frac{1}{π ^{2} n ^{2}} \cdot \int_{0}^{nπ 4} u cos (\frac{u}{4}) d u

Apply Integration By Parts

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

= \frac{1}{π ^{2} n ^{2}} [4 u sin (\frac{u}{4}) - \int 4 sin (\frac{u}{4}) d u]_{0}^{4 πn}

\int 4 sin (\frac{u}{4}) d u = - 16 cos (\frac{u}{4})

= \frac{1}{π ^{2} n ^{2}} [4 u sin (\frac{u}{4}) - (- 16 cos (\frac{u}{4}))]_{0}^{4 πn}

Simplify

= \frac{1}{π ^{2} n ^{2}} [4 u sin (\frac{u}{4}) + 16 cos (\frac{u}{4})]_{0}^{4 πn}

Compute the boundaries:

16 (- 1)^{n} - 16

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

Simplify

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

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

\int_{t}^{3 t} (\frac{sin ( x )}{3 x}) d x

\int_{\frac{π}{3}}^{\frac{π}{2}} csc^{2} (\frac{x}{2}) d x

\int_{0}^{1} \frac{x ( 1 + 3 y ^{2} )}{4} d y

\int_{0}^{x} sin (x - t) d t

What is the integral from 0 to 4 of xcos((npix)/4) ?
The integral from 0 to 4 of xcos((npix)/4) is (16(-1)^n-16)/(pi^2n^2)