Aus Online Mathematik Brückenkurs 2

we look at the formula for partial integration,

\displaystyle \int{f\left( x \right)}g\left( x \right)\,dx=F\left( x \right)g\left( x \right)-\int{F\left( x \right){g}'\left( x \right)\,dx}

we see that if we choose \displaystyle f\left( x \right)=\text{sin }x\text{ } and \displaystyle g\left( x \right)=x+\text{1}, then the factor \displaystyle g\left( x \right) will be differentiated to a constant on the right-hand side of the integral. Naturally, this presupposes that we can find a primitive function for \displaystyle f\left( x \right) (which we can) and that we can then integrate it. Let's try!

\displaystyle \begin{align} & \int{\left( x+1 \right)}\sin x\,dx=\left( x+1 \right)\centerdot \left( -\cos x \right)-\int{1\centerdot }\left( -\cos x \right)\,dx \\ & =-\left( x+1 \right)\cos x+\int{\cos x}\,dx \\ & =-\left( x+1 \right)\cos x+\sin x+C \\ \end{align}