What is the antiderivative of $(x - 2) sin x$ $(x-2)sinx$ ?

Question

1306 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-17T06:53:09

$= - (x - 2) cos x + sin x + C$ $= - (x-2) cos x + sin x + C$

do an IBP

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

$= \int (x - 2) \frac{d}{d x} (- cos x) d x + C$ $= int (x-2) d/dx (- cos x) dx + C$

$= (x - 2) (- cos x) - \int \frac{d}{d x} (x - 2) (- cos x) d x + C$ $= (x-2) (- cos x) - int d/dx (x-2) (- cos x) dx + C$

$= - (x - 2) cos x + \int cos x d x + C$ $= - (x-2) cos x + int cos x dx + C$

$= - (x - 2) cos x + sin x + C$ $= - (x-2) cos x + sin x + C$

What is the antiderivative of (x-2)sinx(x−2)sinx(x-2)sinx?