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February 14th, 2017, 03:21 AM  #1 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40  Can someone show me the quickest and easiest way to integrate this..?
Can someone show me the quickest and easiest way to integrate this..? $\displaystyle \int { \sqrt { { a }^{ 2 }{ x }^{ 2 } } } dx$ Please leave your step by step instructions below. Thankyou very much indeed!! 
February 14th, 2017, 04:23 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,982 Thanks: 1575 
The fastest way would be to reference a table of integrals. Look at #8 in the link Table Failing that, I'd use the trig substitution $x=a\sin{t} \implies dx= a\cos{t} \, dt$ $\displaystyle \int \sqrt{a*2  a^2\sin^2{t}} \cdot a\cos{t} \, dt$ $\displaystyle a \int \sqrt{a^2(1\sin^2{t})} \cos{t} \, dt$ $\displaystyle a^2 \int \cos^2{t} \, dt$ use the trig identity $\cos^2{t} = \dfrac{1+\cos(2t)}{2}$ $\displaystyle \dfrac{a^2}{2} \int 1+\cos(2t) \, dt$ $\dfrac{a^2}{2}\bigg[t + \dfrac{\sin(2t)}{2}\bigg]+C$ $x=a\sin{t} \implies t = \arcsin\left(\dfrac{x}{a}\right) \text{ and } \cos{t} = \dfrac{\sqrt{a^2x^2}}{a}$ therefore, $\sin(2t)=2\sin{t}\cos{t}=2 \cdot \dfrac{x}{a} \cdot \dfrac{\sqrt{a^2x^2}}{a} = \dfrac{2x\sqrt{a^2x^2}}{a^2}$ substitution back into the antiderivative ... $\dfrac{a^2}{2}\bigg[\arcsin\left(\dfrac{x}{a}\right) + \dfrac{x\sqrt{a^2x^2}}{a^2} \bigg]+C$ $\dfrac{a^2}{2}\arcsin\left(\dfrac{x}{a}\right)+ \dfrac{x\sqrt{a^2x^2}}{2}+C$ 
February 14th, 2017, 05:43 AM  #3 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40 
I think you complicated the final bit... But nevertheless, you provided the answer I was looking for. Many thanks. 
February 14th, 2017, 06:38 AM  #4  
Math Team Joined: Jul 2011 From: Texas Posts: 2,982 Thanks: 1575  Quote:
Quote:
 
February 19th, 2017, 02:15 AM  #5 
Newbie Joined: Feb 2017 From: India Posts: 4 Thanks: 0  Sent from my Micromax AQ4502 using Tapatalk 
February 19th, 2017, 03:34 AM  #6 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,230 Thanks: 908 Math Focus: Wibbly wobbly timeywimey stuff. 
@Soumyadip Sorry, but I've seen Egyptian hieroglyphs that are easier to read. If someone else can read it then fine but otherwise could you write it out more slowly? It looks to me like your writing is bad due to how fast you are writing. Dan 
February 19th, 2017, 05:32 AM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
It's an approach that works, but whatever method one uses to evaluate $\displaystyle \int \frac{1}{\sqrt{a^2x^2}}\,\mathrm dx$ (table of integrals, trigonometric substitution) works equally well for the original integral, so you've essentially just added a step.

February 19th, 2017, 07:51 AM  #8 
Newbie Joined: Feb 2017 From: India Posts: 4 Thanks: 0 
OK I'm sry for my bad handwriting It's in my habit. U can integrate the function easily by parts Taking sqrt(a^2  x^2) as the 1st part and 1 as the 2nd one.. Sent from my Micromax AQ4502 using Tapatalk 

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