Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 1 / 11 / c

On an interval with $\sin x\ne 0$, divide by $\sin x$ and use integrating factor $1/\sin x$. ThenPut $u=\sin x$ in the integral: $\int {\cos }^{3}xdx/{\sin }^{2}x=\int (u^{-2}-1)du=-u^{-1}-u+C$. Multiplying back givesAlthough the derivation divided by $\sin x$, this final expression is smooth and directly satisfies the original undivided equation at every real $x$, including multiples of $\pi$.