Question

# Evaluate the following definite integral. \int_{0}^{\pi}x\sin xdx

Applications of integrals
Evaluate the following definite integral.
$$\displaystyle{\int_{{{0}}}^{{\pi}}}{x}{\sin{{x}}}{\left.{d}{x}\right.}$$

2021-03-10
$$\displaystyle{\int_{{{0}}}^{{\pi}}}{x}{\sin{{x}}}{\left.{d}{x}\right.}$$
Integrate by parts:
$$\displaystyle\int{f}{g}'={f}{g}-\int{f}'{g}$$
f=x, f'=1
$$\displaystyle{g}'={\sin{{x}}},{g}=-{\cos{{x}}}$$
$$\displaystyle\Rightarrow{{\left[-{x}{\cos{{x}}}\right]}_{{{0}}}^{{\pi}}}-{\int_{{{0}}}^{{\pi}}}-{\cos{{x}}}{\left.{d}{x}\right.}$$
Solve $$\displaystyle{\int_{{{0}}}^{{\pi}}}-{\cos{{x}}}{\left.{d}{x}\right.}$$
Apply linearity:
$$\displaystyle-{\int_{{{0}}}^{{\pi}}}{\cos{{x}}}{\left.{d}{x}\right.}$$
Solve $$\displaystyle{\int_{{{0}}}^{{\pi}}}{\cos{{x}}}{\left.{d}{x}\right.}$$
It is a standard integral:
$$\displaystyle-{\int_{{{0}}}^{{\pi}}}{\cos{{x}}}{\left.{d}{x}\right.}=-{{\left[{\sin{{x}}}\right]}_{{{0}}}^{{\pi}}}$$
Plugging in solved integrals,
$$\displaystyle{{\left[{\sin{{x}}}-{x}{\cos{{x}}}\right]}_{{{0}}}^{{\pi}}}$$
$$\displaystyle={\sin{{\left(\pi\right)}}}-\pi{\cos{{\left(\pi\right)}}}-{\left[{\sin{{\left({0}\right)}}}-{0}{\cos{{\left({0}\right)}}}\right]}$$
$$\displaystyle={0}-\pi{\left(-{1}\right)}-{\left[{0}-{0}\right]}$$
$$\displaystyle=\pi$$