Definite Integration problem I was working on a small Integration problem, where i was needed to solve another integral:
$$I= \int_0^{\pi} \frac{1}{1+\cos^2x} dx $$
Working:
$$\int_0^{\pi} \frac{1}{1+\cos^2x} dx=\int_0^{\pi} \frac{\sec^2x}{\sec^2x+1} dx$$
$$=\int_0^{\pi} \frac{\sec^2x}{2+\tan^2x} dx$$
Substituting, $\tan x=t, \sec^2x dx=dt$
$$\implies \frac{\sec^2x}{2+\tan^2x}dx=\frac{dt}{2+t^2}$$
which gives us $$I= \left[
      \begin{array}{cc|c}{\frac{1}{\sqrt2}\tan^{-1}\frac{\tan x}{\sqrt2}}
      \end{array}
    \right]_0^\pi$$
I think the answer should be zero. After seeing the answer at WolframAlpha, i think i am doing a serious mistake. Ay help will be appreciated.
 A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
I&=\color{#00f}{\large\int_{0}^{\pi}{\dd x \over 1 + \cos^2\pars{x}}}=
2\int_{0}^{\pi/2}{\dd x \over 1 + \sin^{2}\pars{x}}
=
2\int_{0}^{\pi/2}{\dd x \over 1 + \bracks{1 - \cos\pars{2x}}/2}
\\[3mm]&=
2\int_{0}^{\pi}{\dd x \over 3 - \cos\pars{x}}
=2\int_{0}^{\infty}{1 \over 3 - \pars{1 - t^{2}}/\pars{1 + t^{2}}}
\,{2\,\dd t \over 1 + t^{2}}
=4\int_{0}^{\infty}{1 \over 4t^{2} + 2}\,\dd t
\\[3mm]&=\root{2}\int_{0}^{\infty}{1 \over \pars{\root{2}t}^{2} + 1}\,\root{2}\,\dd t
=\color{#00f}{\large{\root{2} \over 2}\,\pi}
\end{align}
where $\ds{t \equiv \tan\pars{x \over 2}}$.
A: Hint:
Is $t = \tan x$ a valid substitution, in the given range: $[0, \pi]$? Is it continuous in that interval?
Read this. That'll show you what's wrong with your method. The correct way is as suggested by Yiorgos.
A: Hint. First observe that
$$I= \int_0^{\pi} \frac{1}{1+\cos^2x} dx =
2\int_0^{\pi/2} \frac{1}{1+\cos^2x} dx,
$$
and then carry out your transformation.
