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I am having this stupid doubt while solving atleast a dozen questions as of now.Let say we have $$\int_0^\pi xdx = \frac{\pi ^2}{2}$$Just to demonstrate lets say tanx=u and substitute for x, the integral then becomes $$\int_0^0 \frac{tan^{-1}{u}}{1+u^2}du$$What am I doing wrong with the limits ? Both $tan0$ and $tan \pi$ are 0 but clearly the new limit evaluates to 0.

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  • $\begingroup$ Sorry I was just testing whether I typed the correct mathjax,didn't expect a quick response as yours $\endgroup$ Feb 21, 2020 at 14:39
  • $\begingroup$ Oh dude! so prompt to put a close vote. $\endgroup$ Feb 21, 2020 at 14:42
  • $\begingroup$ There are lots of similar questions on this site already, see for example this one: math.stackexchange.com/questions/2380669/… $\endgroup$ Feb 21, 2020 at 15:49

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The problem is that $\tan x$ isn't monotonic on $[0,\,\pi]$. If you split the integration range into regions on which it is monotonic, it would work:

$$\int_0^{\pi/2}xdx+\int_{\pi/2}^\pi xdx=\int_0^\infty\frac{\arctan udu}{1+u^2}+\int_{-\infty}^0\frac{(\pi+\arctan u)du}{1+u^2}.$$Note the one-sided approaches of the singularity at $\pi/2$ tend to $\pm\infty$, so the second integral on the right-hand sie isn't $-1$ times the first. Meanwhile, the last integrand's numerator has a complication: for obtuse $x$, $u=\tan x\implies x=\pi+\arctan u$. In fact, $u\mapsto -u$ in the last integral converts its range to $0,\,\infty$, allowing us to sum the two $u$ integrals as$$\int_0^\infty\frac{\arctan u+\pi-\arctan u}{1+u^2}du=\pi\int_0^\infty\frac{du}{1+u^2}=\frac{\pi^2}{2}.$$

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In this case, your substitution is equivalent to $x=\arctan u,$ and as the range of the arctangent is $(-π/2,π/2),$ it follows that you cannot use this substitution for your integral in the range $[π/2,π].$ To do so, you may need to make a preliminary substitution in the second range, something like $x=2y,$ say.

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