Aus Online Mathematik Brückenkurs 2

The integrand consists of two factors, so partial integration is a plausible method. The most obvious thing to do is to choose \displaystyle x^{2} as the factor that we will differentiate and \displaystyle \cos x as the factor that we will integrate. Admittedly, the \displaystyle x^{2} -factor will not be differentiated away, but its exponent decreases by \displaystyle \text{1} and this makes the integral a little easier:

\displaystyle \int{x^{2}\cos x\,dx=x^{2}\centerdot \sin x-\int{2x\centerdot \sin x\,dx}}

We can attack the integral on the right-hand side in the same way. Let \displaystyle 2x be the factor that we differentiate and \displaystyle \sin x the factor that we integrate. This time, we have only one factor left:

\displaystyle \begin{align} & \int{2x\centerdot \sin x\,dx}=2x\centerdot \left( -\cos x \right)-\int{2\centerdot }\left( -\cos x \right)\,dx \\ & =-2x\cos x+2\int{\cos x\,dx} \\ & =-2x\cos x+2\sin x+C \\ \end{align}

All in all, we obtain

\displaystyle \begin{align} & \int{x^{2}\cos x\,dx=x^{2}\centerdot \sin x-\left( -2x\cos x+2\sin x+C \right)} \\ & =x^{2}\centerdot \sin x+2x\cos x-2\sin x+C \\ \end{align}

For more difficult integrals, it is quite normal to have to work step by step before getting the final answer.