Question

Solve
$f\left(x\right)={\int }_0^{2}t(\sin x-\sin t)dt$

Answer 1

Given,

$f\left(x\right)={\int }_0^{2}t(\sin \left(x\right)-\sin \left(t\right))dt$

apply integration by parts,

$\int u.vdx=u\int vdx$ $-\int (\frac{du}{dx}\int vdx)dx$

Here, $u=(\sin \left(x\right)-\sin \left(t\right)),v=t$

$f\left(x\right)={[\frac{1}{2}{t}^2(\sin \left(x\right)-\sin \left(t\right))-\int -\frac{1}{2}{t}^2\cos \left(t\right)dt]}_0^2$

$={[\frac{1}{2}{t}^2(\sin \left(x\right)-\sin \left(t\right))-(-\frac{1}{2}{t}^2\sin \left(t\right)-t\cos \left(t\right)+\sin \left(t\right))]}_0^2$

$={[\frac{1}{2}{t}^2\sin \left(x\right)+t\cos \left(t\right)-\sin \left(t\right)]}_0^2$

$=2\sin \left(x\right)+2\cos \left(2\right)-\sin \left(2\right)$