What is the integral from 0 to pi of x(-cos(x)+1) ?

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

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integral from 0 to pi of x(-cos(x)+1)

\int_{0}^{π} x (- cos (x) + 1) d x

\frac{π ^{2} + 2}{2} + 1

Decimal

6.93480 \dots

Solution steps

\int_{0}^{π} x (- cos (x) + 1) d x

Apply Integration By Parts

: [x (- sin (x) + x) - \int - sin (x) + x d x]_{0}^{π}

= [x (- sin (x) + x) - \int - sin (x) + x d x]_{0}^{π}

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

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

Simplify

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

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

Compute the boundaries:

\frac{π ^{2} + 2}{2} + 1

= \frac{π ^{2} + 2}{2} + 1

\int_{0}^{2 π} \frac{π - x}{2} d x

\int_{1}^{2} (7 x^{6} - 3 x^{2} + 2 x) d x

\int_{- π}^{π} x sin (2 x) d x

\int_{1}^{4} 5 x^{2} d x

\int_{0}^{0.5} \frac{sin ( 3 x )}{x} d x

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