Answer
$$\int\Big(\frac{1}{\sqrt{2\theta-\pi}}+2\sec^2(2\theta-\pi)\Big)d\theta=\sqrt{2\theta-\pi}+\tan(2\theta-\pi)+C$$
Work Step by Step
$$A=\int\Big(\frac{1}{\sqrt{2\theta-\pi}}+2\sec^2(2\theta-\pi)\Big)d\theta$$ $$A=\int((2\theta-\pi)^{-1/2})+2\sec^2(2\theta-\pi))d\theta$$
We set $a=2\theta-\pi$, then $$da=2d\theta$$ $$d\theta=\frac{1}{2}da$$
Therefore, $$A=\frac{1}{2}\int (a^{-1/2}+2\sec^2 a)da$$ $$A=\frac{1}{2}\Big(\int a^{-1/2}da+2\int\sec^2 ada\Big)$$ $$A=\frac{1}{2}\Big(\frac{a^{1/2}}{\frac{1}{2}}+2\tan a\Big)+C$$ $$A=\sqrt a+\tan a+C$$ $$A=\sqrt{2\theta-\pi}+\tan(2\theta-\pi)+C$$
