$\int\sqrt{1+\cos(2ax)}\,dx$

Using \(1+\cos(2u)=2\cos^2u\), \[ \sqrt{1+\cos(2ax)}=\sqrt2\,|\cos(ax)|. \] This equals \(\sqrt2\cos(ax)\) only on intervals where \(\cos(ax)\ge0\).

If \(a\ne0\), on any interval on which \(\cos(ax)\) has constant sign, \[ \int \sqrt{1+\cos(2ax)}\,dx =\begin{cases} \dfrac{\sqrt2}{a}\sin(ax)+C,&\cos(ax)\ge0,\\[6pt] -\dfrac{\sqrt2}{a}\sin(ax)+C,&\cos(ax)\le0. \end{cases} \] These local primitives glue continuously across the zeros of \(\cos(ax)\) after adjusting the constants. Equivalently, a global primitive is \[ \frac{\sqrt2}{a}H(ax)+C, \] where, for \[ k=\left\lfloor\frac{u+\pi/2}{\pi}\right\rfloor, \qquad H(u)=2k+\sin(u-k\pi). \] Then \(H'(u)=|\cos u|\) wherever differentiated, including across the joining points by continuity of the derivative. If \(a=0\), the integrand is the constant \(\sqrt2\) and the primitive is \(\sqrt2 x+C\).