Aus Online Mathematik Brückenkurs 2

The integrand consists of two factors, so integration by parts 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 1 and this makes the integral a little easier,

\displaystyle \int x^2\cdot\cos x\,dx = x^2\cdot\sin x - \int 2x\cdot\sin x\,dx\,\textrm{.}

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\cdot \sin x\,dx &= 2x\cdot (-\cos x) - \int 2\cdot (-\cos x)\,dx\\[5pt] &= -2x\cos x + 2\int \cos x\,dx\\[5pt] &= -2x\cos x + 2\sin x + C\,\textrm{.} \end{align}

All in all, we obtain

\displaystyle \begin{align} \int x^2\cos x\,dx &= x^2\cdot\sin x - (-2x\cos x+2\sin x+C)\\[5pt] &= x^2\sin x + 2x\cos x - 2\sin x + C\,\textrm{.} \end{align}

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