How do I integrate $\sqrt{1+\sec2x}$?

Question

I tried converting $\sec 2x$ into $1/\cos 2x$, and then performing further simplification which led me to $$\int \sqrt{\frac2{1-\tan^2 x}} dx$$

I'm unable to take it from here.

Answer 1

$$I=\int\sqrt{1+\sec2x}\ \mathrm dx$$ As has been already been noted, $$\sqrt{1+\sec2x}=\frac{\sqrt2|\cos x|}{\sqrt{1-2\sin^2x}}$$ Hence we substitute $t=\sqrt2\sin x$, $$I=\int\frac{\mathrm dt}{\sqrt{1-t^2}}$$ We now substitute $t=\sin u$: $$I=\int\frac{\cos u}{\sqrt{1-\sin^2u}}\mathrm du$$ $$I=\int\frac{\cos u}{\sqrt{\cos^2u}}\mathrm du$$ $$I=\int\mathrm du$$ $$I=u$$ $$I=\arcsin t$$ $$I=\arcsin\big(\sqrt2\sin x\big)+C$$

Answer 2

It is $$\cos(2x)=2\cos^2(x)-1$$ so you will get $$\frac{2\cos^2(x)}{2\cos^2(x)-1}$$,now use the tan-half angle substitution.

Answer 3

Hint:

$$\sqrt{1+\sec2x}=\dfrac{\sqrt2|\cos x|}{\sqrt{1-2\sin^2x}}$$

Substitute $\sqrt2\sin x=y$