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

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.

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  • $\begingroup$ The identity $1+\tan^2 t=\sec^2t$ might be useful, you could certainly try a substitution with $u=\cos t$ at that point... $\endgroup$
    – abiessu
    Oct 8 '17 at 1:37
  • $\begingroup$ What is the source of this problem? $\endgroup$ Oct 8 '17 at 2:23
  • $\begingroup$ Hint:Use mathworld.wolfram.com/WeierstrassSubstitution.html $\endgroup$ Oct 8 '17 at 2:52
  • $\begingroup$ @CarlSchildkraut Find the length of the curve: $r(t) = \cos t \;\mathbf i + \sin t \;\mathbf j + \ln{ \cos t} \;\mathbf k, 0 \le t \le \frac{\pi}{4}$ $\endgroup$ Oct 8 '17 at 3:12
  • $\begingroup$ @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. $\endgroup$ Oct 8 '17 at 3:14
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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.

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