integral

[slender ridge]

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[left lynx]

veganmf

veganmf

veganmf

[left hinge]

[lost plover]

multiply integrand into (sinx)^2 -xsin(x) and separate to

$\int \sin^2(x)dx-\int x\sin(x)dx$

[left lynx]

imabot01

[lost plover]

use integration by parts on the xsin(x) and transform the (sinx)^2 by double angle for cos then integrate

apply the limits of integration when done

the answer is like

$\frac{1}{2}x - \frac{1}{4}\sin(2x) + x\cos(x) - \sin(x) + C]_{0}^{2022\pi}$

[left lynx]

imabot01

[lost plover]

i think

[blazing axle]

no it's sin(sin(x) - x), not sin(x)(sin(x) - x)

if you graph the function you can find a very nice symmetry

[lost plover]

errrr

it just equal zero then i think

[fervent iris]

Hint: sin(sin(x) - x) is π-antiperiodic.

[fervent iris]

A function f(x) is P-antiperiodic if f(x + P) = -f(x). There are two consequences of that:

1. f(x) is 2P-periodic.
2. If f(x) is integrable over any interval of length equal to its period, that integral will be 0.