My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree8Thanks
Reply
 
LinkBack Thread Tools Display Modes
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!
CameronBerryPhysics is offline  
 
December 8th, 2014, 07:43 AM   #2
Senior Member
 
Joined: Apr 2014
From: Glasgow

Posts: 2,156
Thanks: 731

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
Benit13 is offline  
December 8th, 2014, 08:43 AM   #3
Math Team
 
Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

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:

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

Thanks from topsquark and szz
soroban is offline  
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
CameronBerryPhysics is offline  
December 8th, 2014, 09:20 AM   #5
Math Team
 
skeeter's Avatar
 
Joined: Jul 2011
From: Texas

Posts: 2,947
Thanks: 1555

Quote:
Originally Posted by CameronBerryPhysics View Post
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
your link requires a username & password ...
skeeter is offline  
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)
Attached Images
File Type: jpg here.jpg (14.9 KB, 15 views)
CameronBerryPhysics is offline  
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.
Azzajazz is offline  
December 8th, 2014, 05:05 PM   #8
Math Team
 
Joined: Nov 2014
From: Australia

Posts: 689
Thanks: 244

Quote:
Originally Posted by Azzajazz View Post
$dx = 4\cos 4t\,du$
It should be $dx = 4\cos 4t\,dt$
Azzajazz is offline  
December 8th, 2014, 05:39 PM   #9
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,944
Thanks: 1135

Math Focus: Elementary mathematics and beyond
Nice work, Azzajazz.
greg1313 is offline  
December 8th, 2014, 06:13 PM   #10
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,944
Thanks: 1135

Math Focus: Elementary mathematics and beyond
Quote:
Originally Posted by Azzajazz View Post
It should be $dx = 4\cos 4t\,dt$
Edited accordingly.
greg1313 is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
integration, manage, question, substitution



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Integration by substitution question Maths Person Calculus 3 November 29th, 2014 04:57 AM
Is the integration by u substitution always possible szz Calculus 4 November 9th, 2014 01:15 AM
Integration by Substitution bilano99 Calculus 3 November 10th, 2012 10:58 AM
U substitution integration mathman2 Calculus 4 January 5th, 2010 04:31 PM





Copyright © 2019 My Math Forum. All rights reserved.