Integral of $\sqrt {-\sin^2 t + \cos^2 t - \tan^2 t}$

$$\int{\sqrt {(-\sin^2 t + \cos^2 t - \tan^2 t)}}~\textrm{d}t$$

I'm aware of a few trig identities, such as ${\cos^2 t - \sin^2 t} = \cos (2t)$ and $\tan^2 t = \frac{\sin^2 t}{\cos^2 t}$ but these don't seem to help simplify the problem.

No simple $u$-substitution seems to prevent itself, and my attempt to integrate by parts has resulted in an even more difficult integrand.

WolframAlpha and a few different integral calculators cannot seem to solve this.

• The identity $1+\tan^2 t=\sec^2t$ might be useful, you could certainly try a substitution with $u=\cos t$ at that point... Oct 8 '17 at 1:37
• What is the source of this problem? Oct 8 '17 at 2:23
• Oct 8 '17 at 2:52
• @CarlSchildkraut Find the length of the curve: $r(t) = \cos t \;\mathbf i + \sin t \;\mathbf j + \ln{ \cos t} \;\mathbf k, 0 \le t \le \frac{\pi}{4}$ Oct 8 '17 at 3:12
• @alphanumeric0: where the minus sign before $\sin^2 t$ comes from, in such a case? You actually have to integrate $\sqrt{1+\tan^2 t}\,dt$, that is way easier than what you asked. Oct 8 '17 at 3:14

By enforcing the substitution $t=\arcsin u$ we are left with $$\int\sqrt{1-4u^2+2u^4}\frac{du}{1-u^2}=\int\sqrt{2-\frac{1}{(1-u^2)^2}}\,du$$ or, by setting $\frac{1}{1-u^2}=v$, $$\int \frac{\sqrt{2-v^2}}{2v^{3/2}\sqrt{v-1}}\,dv$$ which boils down to an elliptic integral. So, no simple answer to this question.