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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.

MathGeek
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3 Answers3

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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 $\ds{t \equiv \tan\pars{x \over 2}}$.

Felix Marin
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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.

Parth Thakkar
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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.