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Consider the integral $\int_{0}^{\pi/2}{\frac{dt}{1+\cos^{2}(t)}}$.

I want to prove it is equal to $\int_{0}^{+\infty}{\frac{dv}{(1+v^2)(1+\frac{1}{1+v^2})}}=\int_{0}^{+\infty}{\frac{dv}{2+v^{2}}}$

Do you see how to do that ? Is it by making the substitution $v=\tan{t}$ ? Can you help me to understand how it works by showing me each step ?

Edit : how would you calculate $\int_{0}^{\pi}{\frac{dt}{1+\cos^{2}(t)}}$ ?

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up vote 2 down vote accepted

$$\displaystyle \int_{0}^{\pi/2}{\frac{dt}{1+\cos^{2}(t)}}\tag{1}$$

Indeed, let $v = \tan t,\quad\text{ then}\; dv = \sec ^2t\,dt\;\implies\; (1+v^2)dt=dv\;\implies\; dt=\dfrac{dv}{(1+v^2)}$

$\cos^2 t=\dfrac{1}{\sec^2 t}\;=\;\dfrac{1}{1+v^2}$

Substituting, temporarily working with an indefinite integral:

$$\int{\frac{dt}{1+\cos^{2}(t)}}\;=\;\int\frac{dv}{(1+v^2)\cdot(1+\large\frac{1}{(1+v^2)})}\;$$ $$=\;\int\frac{dv}{2+v^2} \; = \;\frac 12 \int \frac{dv}{1 + \large\frac{v^2}{2}} \;=\;\frac 12 \int \frac{dv}{1 + \left(\large\frac{v}{\sqrt{2}}\right)^2}\tag{2}$$

Now, let $\;u= \dfrac{v}{\sqrt{2}}.\;\;$ Then $du = \dfrac{1}{\sqrt{2}}\,dv \implies dv = \sqrt{2}\,du.\;\;$ Then $(2)$ becomes:

$$\frac{\sqrt{2}}{2}\int \frac{du}{1 + u^2} = \frac{\sqrt 2}{2}\tan^{-1}u + C = \frac{1}{\sqrt 2} \tan^{-1}u + C$$

Now we simply back substitute: $u = \dfrac{v}{\sqrt 2}$, so our integral in terms of $v$ is $$\frac 1{\sqrt 2}\tan^{-1}\left(\frac v{\sqrt 2}\right) + C$$ and $v = \tan x$, so evaluating the integral in terms of $x$ with the original bounds of integration gives us: $$\frac1{\sqrt 2}\tan^{-1}\left(\frac{\tan x}{\sqrt 2}\right) + C\;\;= \;\;\frac1{\sqrt 2}\frac x{\tan^{-1}(\sqrt 2)}\,\Big|_0^{\pi/2}$$

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Yes, thank you very much ! You gave the same answer as Abhra but he answered a few minutes before you, so I'll choose his answer. But still thank you very much. – Owl Feb 13 '13 at 15:06
How would you calculate $\int_{0}^{\pi}{\frac{dt}{1+\cos^{2}(t)}}$ ? – Owl Feb 13 '13 at 15:07
If you edit your answer, I'll vote for you :) – Owl Feb 13 '13 at 15:21
Thank you so much ! – Owl Feb 13 '13 at 15:45
It does ! Thank you very much ! I had changed the bounds of the integral but it's ok :) – Owl Feb 13 '13 at 17:35

Putting $u=\tan t$ we have,

$\sec ^2tdt=du$

$\Rightarrow (1+u^2)dt=du$

$\Rightarrow dt=\frac{du}{(1+u^2)}$

$\cos^2 t=1/\sec^2 t=1/(1+u^2)$

Substituting the values in the original integral we have ,




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Thank you very much ! – Owl Feb 13 '13 at 14:31

You can also try the following

$$v=\tan\frac{t}{2}\Longrightarrow \cos t=\frac{1-v^2}{1+v^2}\;\;,\;\;dt=\frac{2}{1+v^2}dv$$



No need to work with an improper integral...

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Thank you very much. How do you find this expression for $\cos{(t)}$ ? – Owl Feb 13 '13 at 14:30
$$\cos t=\cos\left(2\frac{t}{2}\right)=\cos^2\frac{t}{2}-\sin^2\frac{t}{2}=\frac{\cos^2‌​\frac{t}{2}-\sin^2\frac{t}{2}}{\cos^2\frac{t}{2}+\sin^2\frac{t}{2}}=\frac{1-\tan^‌​2\frac{t}{2}}{1+\tan^2\frac{t}{2}}=\frac{1-v^2}{1+v^2}$$ – DonAntonio Feb 13 '13 at 14:40
Thank you very much ! So nice ! How do you integrate $\int\limits_0^1\frac{1+v^2}{1+v^4}dt$ ? – Owl Feb 13 '13 at 19:17
Well, you can try partial fractions: $$v^4+1=\left(v^2+\sqrt 2\, v+1\right)\left(v^2-\sqrt 2 \,v+1\right)$$ – DonAntonio Feb 13 '13 at 19:53
Thank you Don Antonio. – Owl Feb 13 '13 at 20:35

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