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$$I=\int_0^\pi{\frac{dx}{a^2\sin^2x+b^2\cos^2x}}$$ I am unable to understand why a substitution of mine in this problem gives a wrong answer. Here's what I did! $$I=\int_{\tan0}^{\tan\pi}{\frac{d(\tan x)}{a^2\tan^2x+b^2}}=\int_0^0{\frac{dt}{a^2t^2+b^2}}$$ which evaluates to $0$. However if we simply split the integral into two, the solution won't be zero anymore. Something like this: $$I=2\int_0^{\pi/2}{\frac{dx}{a^2\sin^2x+b^2\cos^2x}}$$ because the integrand is evenly symmetric at $x=\pi/2$. Now the integral would result into $$2\int_{\tan0}^{\tan\pi/2}{\frac{d(\tan x)}{a^2\tan^2x+b^2}}=\frac{2}{ab}\biggl(\arctan\bigg(\frac{bt}{a}\bigg)\biggr)_0^\infty=\frac{\pi}{ab}$$ Why is it totally different from the above solution? Also, how do I avoid this from happening in future problems related to definite integration?

Utkarsh Verma
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    You are ignoring the pole of tan in the regime $[0,\pi]$. –  Sep 29 '18 at 23:45
  • As an extreme example take $a=b$. Then you are integrating $\frac{dx}{a^2}$ which valuates to $\frac{\pi}{a^2}$ – AHusain Sep 29 '18 at 23:48
  • @marmot Could you please explain that. – Utkarsh Verma Sep 29 '18 at 23:49
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    You have multiplied top and bottom by $\sec^2 x$. At $x=\pi/2$, that is a priori problematic... – peter a g Sep 29 '18 at 23:55
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    Utkarsh in simple terms $\tan x$ is not continuous in the interval $(0,\pi)$, bcoz it is discontinuous at $\pi/2$, therefore $tan x=t$ is "bad" substitution in the interval $(0,\pi )$. Check out a similar post of mine where I did the exactly same mistake https://math.stackexchange.com/questions/2934074/multiple-answers-in-evaluation-of-a-definite-integral –  Sep 30 '18 at 05:54

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\begin{align} & \int_0^\pi \cdots\, dx = \int_0^{\pi/2} \cdots \, dx + \int_{\pi/2}^\pi \cdots\,dx \\[10pt] = {} & \frac 1 {b^2} \int_0^\infty \cdots\, dt + \frac 1 {b^2} \int_{-\infty}^0 \cdots \, dt \tag 1 \\[10pt] = {} & \frac 1 {b^2} \int_{-\infty}^\infty \cdots \, dt. \end{align}

In line $(1),$ you have $\displaystyle \int_0^{\text{something}} + \int_{\text{something}}^0,$ but for present purposes $+\infty$ and $-\infty$ differ from each other.

Michael Hardy
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