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(1x) + 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. $\displaystyle \frac{1}{\sqrt{1+x^2} + \sqrt{1x^2} + 2}$ instead of $\displaystyle \frac{1}{\sqrt{1+x} + \sqrt{1x} + 2}$ 2. Use a substitution $\displaystyle x = \sin^2(4u)$ instead of $\displaystyle x = \sin(4u)$ Otherwise, it doesn't seem right to me. 
December 8th, 2014, 08:43 AM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, CameronBerryPhysics! Quote:
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: $\:\sqrt{1x} \:=\:\sqrt{1  \cos2\theta} \:=\:\sqrt{2\sin^2\theta} \:=\: \sqrt{2}\sin\theta$ 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  
Math Team Joined: Jul 2011 From: Texas Posts: 2,838 Thanks: 1481  Quote:
 
December 8th, 2014, 11:18 AM  #6 
Newbie Joined: Dec 2014 From: Edinburgh Posts: 5 Thanks: 0 
Ah okay, I've attached a print screen of it to this post (not great quality sorry)

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. Last edited by greg1313; December 8th, 2014 at 06:12 PM. 
December 8th, 2014, 05:05 PM  #8 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244  
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 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,923 Thanks: 1123 Math Focus: Elementary mathematics and beyond  

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