My Math Forum Integration by Substitution question I can't manage

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 December 8th, 2014, 06:41 AM #1 Newbie   Joined: Dec 2014 From: Edinburgh Posts: 5 Thanks: 0 Integration by Substitution question I can't manage Hello there, No idea how to go about this question particularly even though it contains several hints (also not sure how to post this question properly so please help). Show that the Integral with regards to x from 1 to -1 of : 1/[sqrt(1+x)+sqrt(1-x) + 2] using the substitution x = sin4x, is equivalent to the integral with regards to t from pi/8 to 0 of 2cos(4t)/cos^2(t) and a hint is sin^2(2t) + cos^2(2t) = 1 Help would be much appreciated!
 December 8th, 2014, 07:43 AM #2 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,142 Thanks: 726 Math Focus: Physics, mathematical modelling, numerical and computational solutions Can you please check which of the following is true: 1. $\displaystyle \frac{1}{\sqrt{1+x^2} + \sqrt{1-x^2} + 2}$ instead of $\displaystyle \frac{1}{\sqrt{1+x} + \sqrt{1-x} + 2}$ 2. Use a substitution $\displaystyle x = \sin^2(4u)$ instead of $\displaystyle x = \sin(4u)$ Otherwise, it doesn't seem right to me. Thanks from topsquark and szz
December 8th, 2014, 08:43 AM   #3
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Hello, CameronBerryPhysics!

Quote:
 $\displaystyle \int^1_{-1}\frac{dx}{\sqrt{1+x}+\sqrt{1-x} + 2}$

I don't understand their substitution.

$\sqrt{1+\sin4x}\,$ and $\,\sqrt{1-\sin4x}\,$ do not simplify to anything useful.

I would use: $\:x = \cos2\theta \quad\Rightarrow\quad dx \:=\:-2\sin2\theta\,d\theta \:=\:-4\sin\theta\cos\theta\,d\theta$

Then: $\:\sqrt{1+x} \:=\:\sqrt{1 + \cos2\theta} \:=\:\sqrt{2\cos^2\theta} \:=\:\sqrt{2}\cos\theta$

And: $\:\sqrt{1-x} \:=\:\sqrt{1 - \cos2\theta} \:=\:\sqrt{2\sin^2\theta} \:=\: \sqrt{2}\sin\theta$

But then I'm not sure how to integrate:

$\quad\displaystyle \int^{\pi}_{\frac{\pi}{2}} \frac{-4\sin\theta\cos\theta\,d\theta}{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta + 2}$

 December 8th, 2014, 09:16 AM #4 Newbie   Joined: Dec 2014 From: Edinburgh Posts: 5 Thanks: 0 The question is part B.4. in this paper. I can do parts i), ii) and iv). https://exampapers.ed.ac.uk/record/4...+for+physics+1
December 8th, 2014, 09:20 AM   #5
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Quote:
 Originally Posted by CameronBerryPhysics The question is part B.4. in this paper. I can do parts i), ii) and iv). https://exampapers.ed.ac.uk/record/4...+for+physics+1

December 8th, 2014, 11:18 AM   #6
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Ah okay, I've attached a print screen of it to this post (not great quality sorry)
Attached Images
 here.jpg (14.9 KB, 15 views)

 December 8th, 2014, 05:02 PM #7 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 I like this question! Let$$I = \int_{-1}^1 \,\dfrac{1}{\sqrt{1 + x} + \sqrt{1 - x} + 2}\,dx$$ And note that the integrand is even. Thus$$I = 2\int_0^1 \,\dfrac{1}{\sqrt{1 + x} + \sqrt{1 - x} + 2}\,dx$$ Use the substitution$$x = \sin 4t \\ dx = 4\cos 4t\,dt$$ Then$$I = 2\int^{\pi / 8}_0\,\dfrac{4\cos 4t}{\sqrt{1 + \sin 4t} + \sqrt{1 - \sin 4t} + 2}\,dt$$ Now, note that \begin{align*}1 + \sin 4t &= \cos^22t + \sin^22t + 2\cos 2t\sin 2t \\ &= (\cos 2t + \sin 2t)^2 \end{align*} and\begin{align*} 1 - \sin 4t &= \cos^22t + \sin^22t - 2\cos 2t\sin 2t \\ &= (\cos 2t - \sin 2t)^2 \end{align*} So\begin{align*} I &= 2\int^{\pi / 8}_0\,\dfrac{4\cos 4t}{\cos 2t + \sin 2t + \cos 2t - \sin 2t + 2}\, dt \\\\ &= 2\int^{\pi / 8}_0\,\dfrac{4\cos 4t}{2\cos 2t + 2}\,dt \\\\ &= 2\int^{\pi / 8}_0\,\dfrac{2\cos 4t}{\cos 2t + 1}\,dt \\\\ &= 2\int^{\pi / 8}_0\,\dfrac{2\cos 4t}{2\cos^2t}\,dt \\\\ &= \int^{\pi / 8}_0\,\dfrac{2\cos 4t}{\cos^2t}\,dt \end{align*} You should be able to do the next part with all your previous answers. Thanks from greg1313, topsquark, Benit13 and 1 others Last edited by greg1313; December 8th, 2014 at 06:12 PM.
December 8th, 2014, 05:05 PM   #8
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Quote:
 Originally Posted by Azzajazz $dx = 4\cos 4t\,du$
It should be $dx = 4\cos 4t\,dt$

 December 8th, 2014, 05:39 PM #9 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,923 Thanks: 1123 Math Focus: Elementary mathematics and beyond Nice work, Azzajazz.
December 8th, 2014, 06:13 PM   #10
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Quote:
 Originally Posted by Azzajazz It should be $dx = 4\cos 4t\,dt$
Edited accordingly.

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### integrate of 1/2t cos 4t^2 dt

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