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Evaluate

$$I= \int_0^π \frac{1}{1+2sin^2x}dx$$

Answer Options:

  1. $2$
  2. $\pi$
  3. $\frac{π}{√3}$

I need some suggestion here.I think there some trick involved here. I have tried by dividing by cos^2x but after I get stuck...

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    If you mean $\displaystyle{\int_0^\pi\frac{dx}{1+\sin(2x)}}$, it's a divergent integral – 2017-02-10
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    this integral doesn't converge by improper integral. – 2017-02-10
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    @adren but I want evaluation of this integral – 2017-02-10
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    For some reason the OP rolled back all the MathJax formatting I did. Is it proper practice to just leave it in a broken format or do I undo the rollback? – 2017-02-10
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    Not that it will *matter*. This question will close. – 2017-02-10
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    @Adren I want to evaluate this integral – 2017-02-10
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    @Jones Your desire to integrate the function doesn't mean it converges. Will-power alone will not suffice – 2017-02-10
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    @adren pls help – 2017-02-10
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    @Adren Pls help to evaluate this integral – 2017-02-10
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    @ClaudeLeibovici Pls help to evaluate this integral – 2017-02-10
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    I am afraid that you do not know how we work on this site. A lot of people are ready to **help** provided that you show what you tried, explain where you are stuck. Nobody would do your homework. So, add your work in the post. – 2017-02-10

4 Answers 4

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Hint -

$\int_0^\pi \frac{1}{1+2\sin^2x}dx$

$= \int_0^\pi \frac{1}{\sin^2x + \cos^2x +2\sin^2x}dx$

$= \int_0^\pi \frac{1}{\cos^2x+3\sin^2x}dx$

Now divide numerator and denominator by $\cos^2x$ and put $\tan x = t$.

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    You solved it or not? Because you are seems to be very confused. – 2017-02-11
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Kinda Useful Hint $$\int_0^\pi \frac{1}{1+2\sin^2x}dx\le\int_0^\pi dx$$

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    Why eliminate $\pi/\sqrt{3}$ when that is the answer. – 2017-02-10
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    my upvote is back after your edit. +1 – 2017-02-10
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    I think we still need to eliminate $2$ as an answer. This does not help there. – 2017-02-10
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    @BrevanEllefsen: I don't see how those inequalities could help to *compute* the requested integral – 2017-02-10
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    @ParamanandSingh You are right, I will have to tighten the bounds. I really misread this graph at first... – 2017-02-10
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    @Adren Doesn't have to. Goal is to select one of the three options the OP has by elimination. Problem is that I can't read a graph to save my life right now and the bounds need to be tighter than what I've got – 2017-02-10
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    @Adren: He is trying to eliminate options from the given answers. This technique does work in objective type tests – 2017-02-10
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    @BrevanEllefsen: Ok ! I had in mind the question of the computation and not the elimination of false answers. Apologies ... – 2017-02-10
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    @ParamanandSingh Best I can get by simple means is that the integral is less than $2.19$ or so. It is clear that $$\int_0^\pi \frac{1}{1+2\sin^2x}dx=2\int_0^{\pi/2} \frac{1}{1+2\sin^2x}dx$$ by symmetry and we can approximate this with a rectangle and triangle to get $$2\int_0^{\pi/2} \frac{1}{1+2\sin^2x}dx < 2\left(\frac{2}{3}\frac{\pi}{2}\right)\left(\frac{1}{2}\frac{\pi}{2}\frac{4}{3}\right)=\frac{2\pi^2}{9}$$ – 2017-02-10
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    Its OK. You managed that far just by approximations. This is commendable so my upvote remains with your answer. – 2017-02-10
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    @ParamanandSingh alright, glad to hear it :) I updated my answer title to match the current state of things – 2017-02-10
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Here is a first hint ...

Now that you have modified your question, we have to compute :

$$I=\int_0^\pi\frac{dx}{1+2\sin^2(x)}$$

First linearizing, using $\cos(2x)=1-2\sin^2(x)$, we get :

$$I=\int_0^\pi\frac{dx}{2-\cos(2x)}$$

Now try to change to the variable to $u=2x$ and look at what you get ...

HINT 2

We have now :

$$I=\frac12\int_0^{2\pi}\frac{du}{2-\cos(u)}=\frac12\int_{-\pi}^{\pi}\frac{du}{2-\cos(u)}=\int_0^{\pi}\frac{du}{2-\cos(u)}$$

The first equality is a consequence of the $2\pi$-periodicity of $u\mapsto\frac{1}{2-\cos(u)}$ and the second results from its parity.

Now change again the variable to $t=\tan\left(\frac{u}{2}\right)$

HINT 3

Using this last change of variable, we get :

$$I=\int_0^\infty\frac{\frac{2\,dt}{1+t^2}}{2-\frac{1-t^2}{1+t^2}}=2\int_0^\infty\frac{dt}{1+3t^2}=\frac23\int_0^\infty\frac{dt}{t^2+\frac13}$$

Finally :

$$I=\frac23\sqrt3\left[\arctan\left(t\sqrt3\right)\right]_0^\infty=\boxed{\frac{\pi\sqrt3}{3}}$$

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    What after that after taking u =2x then limit change from 0 to 2π – 2017-02-10
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    If we use $u = 2x$ then $dx = du/2$ and not $2\,du$. – 2017-02-10
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    @ParamanandSingh: That's true ! sorry ... editing ... – 2017-02-10
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Multiply and divide by sec^2 x in Numerator and Denominator, then take tan x as t. DONE !