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

The integral from 0 to pi/2 of x(-cos(x)+1) is (pi^2-4pi)/8+1

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

\int_{0}^{\frac{π}{2}} x (- cos (x) + 1) d x

\frac{π ^{2} - 4 π}{8} + 1

Decimal

0.66290 \dots

Solution steps

\int_{0}^{\frac{π}{2}} x (- cos (x) + 1) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{π ^{2} - 4 π}{8} + 1

= \frac{π ^{2} - 4 π}{8} + 1

\int_{0}^{0.8} \frac{1}{( 1 - x ) ^{2}} d x

\int_{2}^{4} \frac{2 ( 1 + x )}{27} d x

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

\int_{0}^{2} 3 4 x d x

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

What is the integral from 0 to pi/2 of x(-cos(x)+1) ?
The integral from 0 to pi/2 of x(-cos(x)+1) is (pi^2-4pi)/8+1