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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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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?

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    $\begingroup$ Sure ! I now see my mistake. Thank you very much for answer ! $\endgroup$ – Lucian Feb 25 at 20:50
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    $\begingroup$ You are very welcome :) $\endgroup$ – clathratus Feb 25 at 22:47
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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.

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