I calculate $\int \frac{dx}{\sin^2x+1}=\frac{1}{\sqrt{2}}\arctan(\sqrt{2}\tan x)+c.$ And then I want to calculate $$\int_{0}^{\pi} \frac{dx}{\sin^2x+1}$$. But $\tan\pi=\tan0=0$. So it seems that $\int_{0}^{\pi} \frac{dx}{\sin^2x+1}=0$, but it's not true. Where is mistake in my justification ?
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1https://math.stackexchange.com/questions/2229955/problem-with-calculation-this-integral-int-0-pi-fracdx13-sin2x?rq=1 – Olivier Oloa Feb 25 '19 at 20:48
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2The function you found as the integral is discontinuous at $\pi/2$. – MasB Feb 25 '19 at 20:49
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1Also similar: https://math.stackexchange.com/questions/1356523/what-are-the-restrictions-on-using-substitution-in-integration – Hans Lundmark Feb 26 '19 at 07:41
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Hint: $$\int_0^\pi \frac{dx}{1+\sin(x)^2}=2\int_0^{\pi/2}\frac{dx}{1+\sin(x)^2}$$ And $\arctan \infty=\pi/2$. Can you take it from here?
clathratus
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I made the same mistake before. Refer to [Integral][Please identify problem] $\displaystyle\int \cfrac{1}{1+x^4}\>\mathrm{d} x$
The reason of the problem is that $\arctan(\sqrt{2}\tan x)$ has a jump but the integral should be continuous, so in the two branches of the function (splitted by the jump point), you need to pick 2 different constants to make it continuous. It also apply to similar situations.
Lance
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