Find integral of $\int\frac{\tan x}{\sqrt{\cos^4x-1}}$

Question

Question

How do i find the integral- $\int\frac{\tan x}{\sqrt{\cos^4x-1}}$?

I think that I am supposed to convert the denominator into difference of two squares which I tried and got me nowhere as I just get $\int\frac{\tan x}{\sin x*\sqrt{(\cos^2x+1)}}$

Answer 1

In order to have the result real, I will switch the order of the subtraction. This is equivalent to a multiplication by $i$. $$\int \frac{\tan (x)}{\sqrt{1-\cos^4 (x)}} \mathrm{d} x=\int \frac{\tan (x)}{\cos^2(x) \sqrt{\sec^4(x)-1}} \mathrm{d} x=\int \frac{\tan(x) \sec^2 (x)\; \mathrm{d} x}{|\tan(x)| \sqrt{\sec^2(x)+1}}.$$

Using the substitution $u=\tan(x)$ we get $$\int \frac{\text{sgn}(u) \mathrm{d} u}{\sqrt{u^2+2}}=\text{sgn}(u) \, \text{arcsinh} \left( \frac{u}{\sqrt{2}} \right)+C=\text{sgn} \left( \tan(x) \right) \, \text{arcsinh} \left( \frac{\tan(x)}{\sqrt{2}} \right)+C. $$

Answer 2

One possibility is to notice that$$\frac{\tan x}{\sqrt{\cos^4x-1}}=\frac{\sin x\cos x}{\cos^2x\sqrt{(\cos^2x)^2-1}}.$$So, doing $u=\cos^2x$ and $\mathrm du=-2\sin x\cos x\,\mathrm dx$, one gets$$-\frac12\frac{\mathrm du}{u\sqrt{u^2-1}}.$$