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$\displaystyle\int_0^\frac{\pi}{2}\frac{1}{2-\cos x} \, dx$ using the substitution $t=\tan\frac{1}{2}x$

  • $x=2\tan^{-1}t$

  • $\dfrac{dx}{dt}=\dfrac{2}{1+t^2}$

  • $dx=\dfrac{2}{1+t^2}\,dt$

  • $\displaystyle\int_0^1 \left(\frac{1}{2-\cos x}\right)\left(\frac{2}{1+t^2}\right)\,dt$

Is this the right idea? If so what do I do next?

$\displaystyle\int_0^1\left(\frac{1}{2-\frac{1-t^2}{1+t^2}}\right) \,\left(\frac{2}{1+t^2}\right)\, dt$

$\displaystyle\int_0^1\frac{2}{1+3t^2}\,dt$

$=2\left[\frac{\ln(1+3t^2)}{6t}\right]_0^1$

amWhy
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maxmitch
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    Can you express $\cos x$ in terms of $t$ to complete the substitution? – Mark Bennet May 19 '13 at 17:44
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    And also note that you need to attend to the limits of your integral. – Mark Bennet May 19 '13 at 17:46
  • I am not sure how to, no..? – maxmitch May 19 '13 at 17:49
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    The reason that the $t=\tan \frac x 2$ substitution (sometimes called the Weierstrass Substitution) is so useful, is that $\sin x = \cfrac {2t}{1+t^2}$ and $\cos x = \cfrac{1-t^2}{1+t^2}$ - so it reduces trigonometric integrals to integrals of rational functions. The formulae should remind you of Pythagoras. Your first task is to prove these formulae, and then investigate online, so that you have a strong grasp of what is going on here. This simple substitution is a first introduction to some important mathematical ideas. – Mark Bennet May 19 '13 at 18:04
  • @MarkBennet Which important ideas? I only know the substitution as an isolated trick. – Potato May 19 '13 at 18:04
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    Well this substitution parametrizes the unit circle. The idea is considerably generalised in the study of algebraic curves in algebraic geometry. – Mark Bennet May 19 '13 at 18:14
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    After substituting, I think you should be at $\int_0^1 \frac{2,dt}{1+3t^2}$. The substitution $\sqrt{3}t=u$ then gets us to the end. – André Nicolas May 19 '13 at 19:21
  • @AméricoTavares Use another substitution?! – maxmitch May 19 '13 at 20:00
  • Your last integral should be as indicated by André Nicolas, $$\frac{1}{2-\dfrac{1-t^{2}}{1+t^{2}}}\frac{2}{1+t^{2}}=\frac{2}{2(1+t^{2})-(1-t^{2})}=\frac{2}{1+3t^{2}}=\frac{2}{1+(\sqrt{3}t)^{2}}$$ – Américo Tavares May 19 '13 at 20:01
  • @AméricoTavares can you not just do what I have done?! You end up with a weird decimal when you do what i have done I don't understand why. – maxmitch May 19 '13 at 20:17
  • I've just pointed out that the integrand should be $2/(1+3t^2)$ and not $2/(1+2t^2)$ as it was before your last edit. The final integral should be $$\int_{0}^{1}\frac{2}{1+3t^{2}}dt=\left. \frac{2}{3}\sqrt{3}\arctan t\sqrt{3}\right\vert _{0}^{1}=\frac{2}{9}\pi \sqrt{3}$$ – Américo Tavares May 19 '13 at 20:31
  • A more thorough derivation of this substitution can be found here. – Mårten W Sep 25 '13 at 07:50

3 Answers3

3

If $\;t = \tan\left(\frac 12 x\right)$, i.e., $\,x = 2\tan^{-1}t,\,$ what should $\dfrac 1{2 - \cos x}$ then be?

We need to replace the function (integrand) of $x$ to one expressed as a function of $t$.

What are the new limits for $\,t\,$ if $\;t = \tan\left(\frac 12 x\right)$?

When $x = 0,\;$ $t = \tan\left(\frac 02\right) = 0$. Okay. But, when $x = \pi/2$, the upper limit of integration needs to be $t = \tan\left(\pi/4\right)$

See Weierstrass Substitution for why $\;\cos x = \dfrac{1-t^2}{1+t^2},\;$ and in general, for the logic of using "$t$-substitution": $t = \tan \frac x2$.

ADDED:

After substituting all of the above, we should have the integrand:

$$\frac{1}{2-\dfrac{1-t^{2}}{1+t^{2}}}\frac{2}{1+t^{2}}=\frac{2}{2(1+t^{2})-(1-t‌​^{2})}=\frac{2}{1+3t^{2}}=\frac{2}{1+(\sqrt{3}t)^{2}}$$

So we have

$$\int_0^1 \frac{2}{1+(\sqrt{3}t)^{2}}\,dt = 2\int_0^1 \frac{1}{1+(\sqrt{3}t)^{2}}\,dt\tag{1}$$

Now, I'm afraid to say, the work isn't done yet. We cannot use $\ln|f(t)|$ where $f(t) =1 + \sqrt{3}t)^{2}$ because we do not have an integrand in the form of $\;\dfrac{f'(t)}{f(t)} \,dx$.

But we're all set up with $(1)$ to use the substitution $$u = \sqrt 3 t.\,\implies du = \sqrt 3 dt \implies dt = \dfrac{1}{\sqrt 3} du$$

Then we have an integrand of the form $$2 \int_0^{\sqrt 3} \dfrac{1}{\sqrt 3} \dfrac{1}{1 + u^2}\,du = \dfrac{2}{\sqrt 3}\int_0^{\sqrt 3} \dfrac{1}{1 + u^2}\,du\tag{2}$$

Now, we recall that $$\int \dfrac {1}{1 + u^2} \,du = \tan^{-1}u + C\tag{$\star$}$$

Can you try and finish it from here? Apply $\star$ to the integral given by $(2)$

$(\star)$ See trigonometric substitution for integrals involving $a^2 + u^2$, where $a$ is a constant. Our integral is of the same form, with $a = 1$:

$$\int \frac{du}{a^2 + u^2} = \frac 1a\tan^{-1}\left(\frac ua\right)\,+ C$$

amWhy
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  • $\int_{0}^1$ --- – maxmitch May 19 '13 at 17:52
  • Yes, indeed. You answered before I finished writing...;-) – amWhy May 19 '13 at 17:55
  • So you multiply out $\int_{0}^\frac{\pi}{2} \left(\frac{1}{2-\cos x}\right)\left(\frac{2}{1+t^2}\right),dt$ and then put the new limits in? and then what do you do about the $cosx$? – maxmitch May 19 '13 at 17:57
  • @maxmitch - as soon as you change $dx$ to $dt$ in the integral, you need to put in the limits for $t$ instead of the ones for $x$. – Mark Bennet May 19 '13 at 18:06
  • Note that for $t = \tan\left(x/2\right)$, we have $;\cos x= \dfrac{1-t^2}{1+t^2}$ – amWhy May 19 '13 at 18:12
  • Sorry stupid mistake! I don't remember being taught that $cosx=\frac{1-t}{1+t}$....! (not to be confused with factorial ;) ) – maxmitch May 19 '13 at 18:13
  • is my update correct? – maxmitch May 19 '13 at 19:02
  • See updated solution. When I used substitution again, I changed to bounds of integration so they apply to $u$, so no need to "back substitute" – amWhy May 19 '13 at 21:01
  • @amWhy: Very nice and detailed answer! Very slow day for me, and I have been making lots of really silly errors! Guess I have lots on my mind with another talk this week! Hope all is well! Regards – Amzoti May 20 '13 at 00:36
  • Thanks, Amzoti! Very slow for me today, as well... :-( – amWhy May 20 '13 at 00:38
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Hint:

When $\tan \dfrac{x}{2}=t$ you have $ \cos x= \dfrac{1-t^2}{1+t^2}$

Inceptio
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0

If $x=2\arctan t$ then $\cos x = \cos\left(2\arctan t\right)$. Use the fact that $\cos(2u)=\cos^2u-\sin^2u$. So you get $$ \cos x = \cos^2\left(\arctan t \right) -\sin^2\left(\arctan t \right) $$

If $\varphi=\arctan t$ then $\dfrac t1=t = \tan\varphi=\dfrac{\text{opposite}}{\text{adjacent}}$, so $\cos\varphi=\dfrac{\text{adjacent}}{\text{hypotenuse}}$. So $\text{opposite}=t$, $\text{adjacent}=1$, and by the Pythagorean theorem $\text{hypotenuse}=\sqrt{1+t^2}$.

Then $\cos=\dfrac{\text{adjacent}}{\text{hypotenuse}}$ and so $$ \cos^2(2\arctan t) = \cos^2 (\cdot)-\sin^2(\cdot) = \left(\frac{1}{\sqrt{1+t^2}}\right)^2 - \left(\text{something similar}\right)^2 = \cdots\cdots $$

Michael Hardy
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