<s>[INST] Integral with square root + Trig $$ \int_0^{\pi} \sqrt{-3 \sin(x) - 3 \cos(x) + 5} ~~dx$$
I tried to use substitution:
$$ u = -3 \sin(x) -3 \cos(x) + 5 \\ du = -3 \cos(x) + 3 \sin(x) $$
But because $ ( \cos(x) )' = - \sin(x)$ it does not work that well (the signs are now flipped...)
I even tried to change the $\cos(x)$ into $\sin( x - \frac{\pi}{2} )$ but it didn't get me very far..
I also tried using this online tool: https://www.integral-calculator.com/
And it couldn't find the anti-derivative, but only gave the answer of:
$$ \approx 5.1363$$
I would appreciate your help, really any way possible to solve this ugly question would be appreciated! Thank you very much!
 [/INST] Similar to @Andrew's comment make
$$\sin(x)+\cos(x)=\sqrt{2} \cos \left(x-\frac{\pi }{4}\right)=\sqrt{2} \left(1-2 \sin ^2\left(\frac{x }{2}-\frac{\pi}{8}\right)\right)$$ This makes
$$5-3\sin(x)-3\cos(x)=(5-3 \sqrt{2})+6 \sqrt{2} \sin ^2\left(\frac{x}{2}-\frac{\pi }{8}\right)$$
$$\sqrt{5-3\sin(x)-3\cos(x)}=a\,\sqrt{1+k\sin ^2\left(\frac{x}{2}-\frac{\pi }{8}\right)}$$ with $$a=\sqrt{5-3 \sqrt{2}}\qquad \text{and}\qquad k=\frac{6(6+5 \sqrt{2})}{7} $$
Now, let $\frac{x}{2}-\frac{\pi }{8}=t$ that is to say $x=2t+\frac{\pi }{4}$ to make the antiderivative
$$I=2a \int  \sqrt{1+k \sin^2 (t)}\,dt =2a\,E(t|-k)$$ where appears the elliptic integral  on fthe first kind.
Back to $x$, $a$ and $k$, using the bounds and simplifying, we the have
$$\color{red}{\int_0^{\pi} \sqrt{5-3 \sin(x) - 3 \cos(x) }\,dx =}$$
$$\color{red}{2 \sqrt{5-3 \sqrt{2}} \left(E\left(\frac{\pi }{8}|-\frac{6\left(6+5
   \sqrt{2}\right)}{7} \right)+E\left(\frac{3 \pi }{8}|-\frac{6\left(6+5
   \sqrt{2}\right)}{7} \right)\right)}$$ which is
$$5.1363328412422184497730638111478712316615873410060$$
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