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Having a brain fart. In trying to find the anti-derivative of $$\int \frac{x}{\sqrt{2}-\sin{2x}}dx$$

First I made the substitution: $$\begin{align} \begin{cases} u&=\sqrt{2}-\sin{2x} \to x=\frac{1}{2} \arcsin{(\sqrt{2}-u)} \\ du &=-2\cos{2x}\,dx \to dx=\frac{1}{-2\sqrt{1-(\sqrt{2}-u)^2}}du\end{cases} \end{align}$$

Thus $$I=\frac{-1}{4}\int \frac{\arcsin(\sqrt{2}-u)}{\sqrt{1-(\sqrt{2}-u)^2}}du$$ And getting gitty that we have a function and it's derivative in the integrand:

$$\begin{align} \begin{cases} v&=\sqrt{2}-u \\ -dv &=du \end{cases} \end{align} \implies I=\frac{+1}{4}\int \frac{\arcsin v}{\sqrt{1-v^2}}dv \implies$$

$$\begin{align} \begin{cases} m&=\arcsin{v} \\ dm &=\frac{1}{\sqrt{1-v^2}} dv \end{cases} \end{align} \implies I=\frac{+1}{4}\int m\cdot dm \implies$$

$$\begin{align} I&=\frac{m^2}{8}+C \\ &=\frac{(\arcsin{v})^2}{8}+C \\ &=\frac{[\arcsin{(\sqrt{2}-u)}]^2}{8}+C \\ &=\frac{(2x)^2}{8}+C=\frac{x^2}{2}+C\end{align}$$

And that ain't right....thus I seek counsel and wisdom, and where's my mistake?

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  • $\begingroup$ Am i blind or you didn't substitute $ u=\sqrt{2}-sin(2x) $ in the denominator? $\endgroup$ – valer Sep 21 '17 at 20:09
  • $\begingroup$ @valer sun of a gun..... thank you $\endgroup$ – AmateurMathPirate Sep 21 '17 at 20:10
  • $\begingroup$ I don't think that expresion has a simple primitive see wolframalpha.com/input/?i=x%2F(sqrt(2)-sin(2x)) $\endgroup$ – valer Sep 21 '17 at 20:13
  • $\begingroup$ @valer perhaps by parts then given your assist... $\endgroup$ – AmateurMathPirate Sep 21 '17 at 20:18
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    $\begingroup$ In general, we have $$\int_0^{k\pi}\frac{x}{\sqrt2-\sin2x}~dx ~=~ T_{4k-1}\cdot\bigg(\frac\pi4\bigg)^2$$ and $$\int_0^{(2k+1)\cdot\tfrac\pi2}\frac{x}{\sqrt2-\sin2x}~dx ~=~ T_{4k+2}\cdot\bigg(\frac\pi4\bigg)^2$$ where $T_n$ represents the n-th triangular number. See also OEIS A$217$. $\endgroup$ – Lucian May 9 '18 at 0:57

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