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What's wrong with the Newton-Leibniz formula? $\int_0^\frac{3\pi}{4}\frac{\sin x}{1+\cos^2x}dx =\arctan(\sec x)|_0^\frac{3\pi}{4} =-\arctan \sqrt{2}-\frac{\pi}{4}$. This gives a wrong answer. Where goes wrong?

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    $\begingroup$ Your antiderivative is wrong. -$\arctan(\cos x)$ is the correct one. $\endgroup$
    – FDP
    May 27 '19 at 7:45
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    $\begingroup$ @FDP The OP's answer also works; differentiate it. $\endgroup$
    – J.G.
    May 27 '19 at 7:47
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    $\begingroup$ @JG: $\frac{3\pi}{4}>\frac{\pi}{2}$ and $\sec x$ doesn't exist for $x=\frac{\pi}{2}$ $\endgroup$
    – FDP
    May 27 '19 at 9:30
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The function $\arctan\sec x$ has a discontinuity at $x=\pi/2$, because $\sec x$ becomes $\infty$ from below and $-\infty$ from above, and its arctangent changes discontinuously from just under $\pi/2$ to just over $-\pi/2$. If you split the integral at $x=\pi/2$, then add the two integrals together, it'll work out.

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$$\int_0^\frac{3\pi}{4}\frac{\sin x}{1+\cos^2x}dx =-\arctan(\cos x)|_0^\frac{3\pi}{4} =-\arctan (-\sqrt{2}/2 )+\frac{\pi}{4} =\arctan (\sqrt{2}/2 )+\frac{\pi}{4}$$

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