Understanding $\int_0^\pi{\frac{dx}{a^2\sin^2x+b^2\cos^2x}}$

Question

$$I=\int_0^\pi{\frac{dx}{a^2\sin^2x+b^2\cos^2x}}$$ 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?

Answer 1

\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.