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integrate sin 2x / √(1 + cos² x) dx

Mathematics
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Hmm, try substitute \[u=\cos^2(x) \\ \\du=-2\cos(x)\sin(x)\] \[-\int\limits \frac{1}{\sqrt{u+1}} \, du\]\ Substitute one more time, \[t=u+1 \\ \\dt=du\] \[-\int\limits \frac{1}{\sqrt{t}} \, dt\] Can you do it now?
Note: from starting, I used sin(2x)=2sin(x)cos(x)
i dont really get it :(

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Thought so, :D I'll rearrange it
in my book it says that the answer is -2 √(1 + cos² x) + c
You have \[\int\limits \frac{\sin 2x }{\sqrt{ \cos ^2x+1}} \, dx\] Change the sin(2x) to 2sin(x)cos(x), That's an identity. \[\int\limits \frac{2sin(x)cos(x) }{\sqrt{ \cos ^2x+1}} \, dx\] Then, let \[u=\cos^2(x) \\ \\du=-2\cos(x)\sin(x) dx \\ \\-du=2\sin(x)\cos(x)dx\] \[\int\limits \frac{-du }{\sqrt{ u+1}} \, \] Substitute one more time, Let \[t=u+1 \\ \\dt=du\] \[\int\limits \frac{-dt }{\sqrt{ t}} \, \] \[-\int\limits \frac{1 }{\sqrt{ t}} dt\, \] \[-\int\limits t^{-\frac{1}{2}}dt\, \] Integrate it, \[-2 \sqrt{t}+c\] Substitute back, \[-2 \sqrt{u+1}+c\] \[-2 \sqrt{cos^2x+1}+c\]
i see... but in my level, we haven't learn how to differentiate cos² x directly, we only learn how to differentiate cos x so i dont really get it is there any other way?
I don't think so, but differentiating cos² x is using chain rule, you should know that before doing integration. Hmm, that's weird.
okay i'll try. thank you!
i only didnt undertsand that part but the rest i do so thanks a lot !

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