What is the integral from 0 to pi of x(pi-x)sin(nx) ?

The integral from 0 to pi of x(pi-x)sin(nx) is -(pi^2(-1)^n)/n-(2(-1)^n-pi^2(-1)^nn^2-2)/(n^3)

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integral from 0 to pi of x(pi-x)sin(nx)

\int_{0}^{π} x (π - x) sin (n x) d x

- \frac{π ^{2} ( - 1 ) ^{n}}{n} - \frac{2 ( - 1 ) ^{n} - π ^{2} ( - 1 ) ^{n} n ^{2} - 2}{n ^{3}}

Solution steps

\int_{0}^{π} x (π - x) sin (n x) d x

Expand

x (π - x) sin (n x) : π x sin (n x) - x^{2} sin (n x)

= \int_{0}^{π} π x sin (n x) - x^{2} sin (n x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{π} π x sin (n x) d x - \int_{0}^{π} x^{2} sin (n x) d x

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

\int_{0}^{π} x^{2} sin (n x) d x = \frac{2 ( - 1 ) ^{n} - π ^{2} ( - 1 ) ^{n} n ^{2} - 2}{n ^{3}}

= - \frac{π ^{2} ( - 1 ) ^{n}}{n} - \frac{2 ( - 1 ) ^{n} - π ^{2} ( - 1 ) ^{n} n ^{2} - 2}{n ^{3}}

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

\int_{0}^{4} ∣ 2 x - 5 ∣ d x

\int_{1}^{s i n (x)} 3 t^{2} d t

\int_{- 1}^{0} x^{5} x^{3} + 1 d x

\int_{- 1}^{1} \frac{1}{1 - x ^{2}} d x

What is the integral from 0 to pi of x(pi-x)sin(nx) ?
The integral from 0 to pi of x(pi-x)sin(nx) is -(pi^2(-1)^n)/n-(2(-1)^n-pi^2(-1)^nn^2-2)/(n^3)