What is the integral from 0 to 3 of xsin((mpix)/3) ?

The integral from 0 to 3 of xsin((mpix)/3) is (9sin(pim)-9pimcos(pim))/(pi^2m^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 3 of xsin((mpix)/3)

\int_{0}^{3} x sin (\frac{mπ x}{3}) d x

\frac{9 sin ( πm ) - 9 πm cos ( πm )}{π ^{2} m ^{2}}

Solution steps

\int_{0}^{3} x sin (\frac{mπ x}{3}) d x

Apply u-substitution

: \int_{0}^{m \cdot 3 π} \frac{u sin ( \frac{u}{3} )}{π ^{2} m ^{2}} d u

= \int_{0}^{m \cdot 3 π} \frac{u sin ( \frac{u}{3} )}{π ^{2} m ^{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} m ^{2}} \cdot \int_{0}^{m \cdot 3 π} u sin (\frac{u}{3}) d u

Apply Integration By Parts

: [- 3 u cos (\frac{u}{3}) - \int - 3 cos (\frac{u}{3}) d u]_{0}^{3 πm}

= \frac{1}{π ^{2} m ^{2}} [- 3 u cos (\frac{u}{3}) - \int - 3 cos (\frac{u}{3}) d u]_{0}^{3 πm}

\int - 3 cos (\frac{u}{3}) d u = - 9 sin (\frac{u}{3})

= \frac{1}{π ^{2} m ^{2}} [- 3 u cos (\frac{u}{3}) - (- 9 sin (\frac{u}{3}))]_{0}^{3 πm}

Simplify

= \frac{1}{π ^{2} m ^{2}} [- 3 u cos (\frac{u}{3}) + 9 sin (\frac{u}{3})]_{0}^{3 πm}

Compute the boundaries:

9 sin (πm) - 9 πm cos (πm)

= \frac{1}{π ^{2} m ^{2}} (9 sin (πm) - 9 πm cos (πm))

Simplify

= \frac{9 sin ( πm ) - 9 πm cos ( πm )}{π ^{2} m ^{2}}

\int_{4}^{6} \frac{1}{x ^{2} - 25} d x

\int_{0}^{\frac{π}{14}} sin^{5} (7 x) d x

\int_{11}^{\infty} \frac{1}{x ^{6}} d x

\int_{π - α}^{2 π} - 1

\int_{0.5}^{1} \frac{3}{4} x (1 - 0.25 x) d x

What is the integral from 0 to 3 of xsin((mpix)/3) ?
The integral from 0 to 3 of xsin((mpix)/3) is (9sin(pim)-9pimcos(pim))/(pi^2m^2)