# Evaluating $\int\limits_0^{\pi} \frac{dx}{1+2\sin^2x}$

After making u substitution two times, I am getting indefinite integral as $$\int\limits\dfrac{dx}{1+2\sin^2x} = \dfrac{\arctan(\sqrt{3}\tan(x))}{\sqrt{3}}+ C$$

I am stuck at working the bounds because I am getting $0$ but the answer needs to be $\pi/\sqrt{3}$. I think $\arctan(x)$ is defined in $-\pi/2 \lt x\lt\pi/2$ while making u substitution. However I don't see how to use this observation to my advantage. Can anybody help in getting the answer and understanding whats going on. Thanks !

• If you indicate what substitutions you made, that will help the diagnosis. Things may become easier for you if you note that by the symmetry of $\sin x$ about $x=\pi/2$, your integral is twice the integral from $0$ to $\pi/2$. Commented Oct 5, 2014 at 7:19
• Yeah I have noticed that from the graph of integrand. I don't think the result of indefinite integral depends on the method I have used... My specific problem was in taking the result from indefinite integral and using it to evaluate definite integral Commented Oct 5, 2014 at 7:22
• Or the indefinite integral makes sense only in (-pi/2 , pi/2) is it ? I am not too sure. The bounds and domain are really confusing :/ Commented Oct 5, 2014 at 7:26
• Evaluating from $0$ to $\pi/2$ and doubling gives $\pi/\sqrt{3}$. Commented Oct 5, 2014 at 7:30
• Commented Dec 1, 2017 at 5:30

You may write \begin{align} \int_0^{\pi/2}\dfrac{dx}{1+2\sin^2x}& =\int_0^{\pi/2}\dfrac{dx}{1+2(1-\cos^2x)}\\\\ &=\int_0^{\pi/2}\dfrac{dx}{3-2\cos^2x}\\\\ &=\int_0^{\pi/2}\dfrac{dx}{3-\frac{2}{1+\tan^2x}}\\\\ &=\int_0^{\pi/2}\dfrac{(1+\tan^2x)}{1+3\tan^2x}dx\\\\ &=\int_0^{\infty}\dfrac{du}{1+3u^2}\\\\ &=\left.\dfrac{\arctan(\sqrt{3}u)}{\sqrt{3}}\right|_{0}^{\infty}\\\\ &=\dfrac{\pi}{2\sqrt{3}} \end{align} and use André Nicolas' comment: your initial integral is twice the preceding.

• Thank you :) Is there a way to work this without using symmetry ? Why splitting the integral 0->pi/2 and pi/2->pi is not working ? Commented Oct 5, 2014 at 7:37
• @rsadhvika It also works with $]\frac{\pi}{2},\pi[$, you have your initial integral $I$ verifying $I=2I_{]0,\frac{\pi}{2}[}=2I_{]\frac{\pi}{2},pi[}$. Commented Oct 5, 2014 at 7:43
• F(pi/2) - F(0) + F(pi) - F(pi/2) gives me F(pi) - F(0) which evaluates to 0. I see I have failed to communicate my specific problem clearly. I'll post a new question... My problem was in using the indefinite integral to evaluate the definite integral... Not in working the indefinite integral. Commented Oct 5, 2014 at 7:45

Your integral is twice the below integral. The reason why your integral is twice the below integral is because it satisfies $f(2a-x)=f(x)$ where $a =\frac{\pi}{2}$.

\begin{align*} \int_{0}^{\pi/2} \frac{1}{1+2\sin^{2}(x)} &= \int_{0}^{\pi/2} \frac{\sec^{2}(x)}{\sec^{2}(x)+2\tan^{2}(x)} \\&= \int_{0}^{\pi/2} \frac{\sec^{2}(x)}{1+3\tan^{2}(x)} \end{align*}

I think that Weierstrass substitution could help. Let $$I=\int \dfrac{dx}{1+2\sin^2x}$$ So, using $t=\tan(\frac{x}{2})$, you arrive to $$I=\int \frac{2 \left(t^2+1\right)}{t^4+10 t^2+1}dt$$ Now, notice that the denominator write $$t^4+10 t^2+1=(t^2+\alpha)(t^2+\beta)$$ with $\alpha=2 \sqrt{6}+5$, $\beta=2 \sqrt{6}-5$. Use partial fraction decomposition and integrate; you should arrive to $$I=\frac{\tan ^{-1}\left(\left(\sqrt{3}-\sqrt{2}\right) t\right)+\tan ^{-1}\left(\left(\sqrt{3}+\sqrt{2}\right) t\right)}{\sqrt{3}}=\frac{\tan ^{-1}\left(\frac{2 \sqrt{3} t}{1-t^2}\right)}{\sqrt{3}}$$ Now, use the bounds for $t$.

We solve the problem in a way which is almost identical to yours, with a small correction. Let $F(x)$ be an antiderivative of our function. Then the integral is $F(\pi)-F(0)$.

The only problem with your solution is that for $\frac{\pi}{2}\lt x\lt \pi$, the function $$\frac{\arctan(\tan(\sqrt{3}x))}{\sqrt{3}}\tag{1}$$ is not the correct antiderivative. The issue is the back substitution.

For recall that $\arctan t$ is the number $t$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ whose tan is $t$. If we want an angle between $\frac{\pi}{2}$ and $\pi$, we need to use $\pi+\arctan t$. In particular, for $x$ near to but below $\pi$, $F(X)$ is given by $$F(x) =\frac{\pi +\arctan(\tan(\sqrt{3}x))}{\sqrt{3}}.\tag{2}$$ Now "plug in" $\pi$ (in the right expression (2) and take away the result of plugging in $0$ (in the right expression (1)). We get $\frac{\pi}{\sqrt{3}}$.

Remark: I would in fact break up the integral into two parts, as a reflexive response to the symmetry. And even if did not, for the substitution I would consider the two intervals separately, since singularities make me nervous. However, that is not necessary. And in the solution above, $\frac{\pi}{2}$ does not even come into the picture.

• André Nicolas, thank you for these additional informations. Commented Oct 5, 2014 at 15:47
• Thank you so much! this cleared up the question that's troubling me :) Commented Oct 8, 2014 at 15:18
• You are welcome. This issue comes up every so often with the inverse trigonometric functions. In principle one should check each time that nothing has gone wrong. In practice, we can usually operate mechanically. But, as your example illustrates, certainly not always. Commented Oct 8, 2014 at 15:25

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} \color{#66f}{\large\int{\dd x \over 1 + 2\sin^{2}\pars{x}}}& =\int{\sec^{2}\pars{x}\,\dd x \over \sec^{2}\pars{x} + 2\tan^{2}\pars{x}} = \overbrace{\int{\sec^{2}\pars{x}\,\dd x \over 3\tan^{2}\pars{x} + 1}} ^{\color{#c00000}{\ds{\mbox{Set}\ t \equiv \tan\pars{x}}}} \\[5mm]&={1 \over \root{3}} \ \underbrace{% \int{\root{3}\,\dd t \over \pars{\root{3}t}^{2} + 1}} _{\color{#c00000}{\ds{\root{3}t \equiv \xi\ \imp\ t = {\root{3} \over 3}\,\xi}}}\ =\ {\root{3} \over 3}\int{\dd\xi \over \xi^{2} + 1} \\[5mm]&={\root{3} \over 3}\,\arctan\pars{\xi}={\root{3} \over 3}\,\arctan\pars{\root{3}t} \end{align}

$$\color{#66f}{\large\int{\dd x \over 1 + 2\sin^{2}\pars{x}} ={\root{3} \over 3}\,\arctan\pars{\root{3}\tan\pars{x}}} + \mbox{a constant}$$