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$$\int \frac{\sin x}{\sqrt{2-\cos^2x}}dx$$

I have tried many trig identities but it seems that none have made a way to the solution.

$$\int \frac{\sin x}{\sqrt{2-\cos^2x}}dx=\int \frac{\sin x}{\sqrt{1+\sin^2x}}dx$$

$$\int \frac{\sin x}{\sqrt{2-\cos^2x}}dx=\int \frac{\sin x}{\sqrt{2+\frac{1+\cos 2x}{2}}}dx=\int \frac{\sin x}{\sqrt{\frac{5+\cos 2x}{2}}}dx$$

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From the very beginning, let $u = \cos x$. Then $du = -\sin x \, dx$, and we have

$$ \int \frac{\sin x}{\sqrt{2 - \cos^2 x}} \, dx = -\int \frac{du}{\sqrt{2 - u^2}}$$

Can you take it from here?

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$$\int \frac { sinx }{ \sqrt { 2-\cos ^{ 2 } x } } dx=-\int { \frac { d\left( \cos { x } \right) }{ \sqrt { 2-\cos ^{ 2 } x } } } $$ and substitute $\\ \cos { x } =t$ so that you will get

$$\\ -\int { \frac { dt }{ \sqrt { 2-{ t }^{ 2 } } } } =-\int { \frac { d\left( \frac { t }{ \sqrt { 2 } } \right) }{ \sqrt { 1-{ \left( \frac { t }{ \sqrt { 2 } } \right) }^{ 2 } } } } =\arccos { \left( \frac { t }{ \sqrt { 2 } } \right) } =\arccos { \left( \frac { \cos { x } }{ \sqrt { 2 } } \right) } +C$$

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  • $\begingroup$ and then $\sqrt{2}\sqrt{1-(\frac{t}{\sqrt{2}})^2}?$ $\endgroup$ – newhere Aug 8 '16 at 19:41
  • $\begingroup$ arccos? shouldn't it be arcsin? $\endgroup$ – newhere Aug 8 '16 at 19:47
  • $\begingroup$ @newhere,no it should be arccos $\endgroup$ – haqnatural Aug 8 '16 at 19:49
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    $\begingroup$ @newhere $\int \frac1{\sqrt{1-t^2}}\,dt=\arcsin x+C$, but $-\int \frac1{\sqrt{1-t^2}}\,dt=\arccos x+C$ (because of the often neglected identity $\arccos x=\frac\pi2-\arcsin x$) $\endgroup$ – user228113 Aug 8 '16 at 20:00
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$$\int \frac{\sin x}{\sqrt{2-\cos^2x}}dx=\int \frac{\sin x}{\sqrt{\frac{3 -\cos 2x}{2}}}dx$$ $$3-\cos 2x=t^2$$ $$\sin x dx=\frac{tdt}{2\cos x}$$ $$3-t^2=2\cos^2 x -1$$ $$\cos x=\sqrt{\frac{4-t^2}{2}}$$ $$\int\frac{dt}{\sqrt{4-t^2}}$$

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