My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree4Thanks
  • 2 Post By skeeter
  • 1 Post By v8archie
  • 1 Post By skipjack
Reply
 
LinkBack Thread Tools Display Modes
May 19th, 2017, 11:06 AM   #1
Member
 
Joined: Jan 2017
From: California

Posts: 77
Thanks: 8

Stuck on this integral

Hi guys.

What is the best way to integrate this?

I can't seem to wrap my head around it.
Attached Images
File Type: gif gif.gif (778 Bytes, 14 views)

Last edited by skipjack; May 19th, 2017 at 11:48 PM.
dthiaw is offline  
 
May 19th, 2017, 11:34 AM   #2
Math Team
 
Joined: Jul 2011
From: Texas

Posts: 2,604
Thanks: 1290

if the integrand is $\left(x + \dfrac{5}{9}\right)^{3/2}$, then the antiderivative is

$\dfrac{2}{5}\left(x + \dfrac{5}{9}\right)^{5/2} + C$


if the integrand is $\left(\dfrac{x + 5}{9}\right)^{3/2} = \dfrac{(x+5)^{3/2}}{27}$, then the antiderivative is

$\dfrac{2}{135}\left(x + 5 \right)^{5/2} + C$
Thanks from v8archie and dthiaw
skeeter is offline  
May 19th, 2017, 11:53 AM   #3
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 6,876
Thanks: 2240

Math Focus: Mainly analysis and algebra
It looks ideal for a simple $u$ substitution.
Thanks from dthiaw
v8archie is offline  
May 20th, 2017, 02:41 AM   #4
Global Moderator
 
Joined: Dec 2006

Posts: 17,722
Thanks: 1359

To try to find $\displaystyle \!\int\! f(x)dx$ using the substitution $x = g(u)$, where $g(u)$ is invertible and differentiable,
find $\dfrac{dx}{du} = g^\prime\!(u)$, then use $\displaystyle \!\int\! f(x)dx = \!\int\! f(g(u))\frac{dx}{du}du = \!\int\! f(g(u))g^\prime\!(u)du$.

After integrating, write the result in terms of $x$ (using the inverse of $g$ if necessary).

Explicitly specifying the domain of $g$ is advisable to assist with that and simplifying $f(g(u))g^\prime\!(u)$.

To find, for example, $\displaystyle \!\int\! \frac{1}{x^2\sqrt{1 - x^2}} dx$, where $0 < x < 1$,
let $x = \cos(u)$, where $0 < u < \pi/2$, so that $\dfrac{dx}{du} = -\sin(u)$ and $\sqrt{1 - x^2} = \sin(u)$.

This gives $\displaystyle \!\int\! \frac{1}{\cos^2(u)\sin(u)} (-\sin(u)) du = \!\int\! -\sec^2(u)du = -\tan(u) + \text{C} = -\frac{\sin(u)}{\cos(u)} + \text{C}$ (where $\text{C}$ is a constant),
which equals $-\dfrac{\sqrt{1 - x^2}}{x} + \text{C}$.
Thanks from dthiaw
skipjack is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
integral, stuck



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Stuck. Please help KaiL Algebra 1 December 7th, 2016 03:07 AM
stuck! natt010 Calculus 4 January 30th, 2016 08:41 AM
I am stuck on an integral problem. CrimeAndPunishment Calculus 1 April 5th, 2014 03:52 PM
Stuck on an improper integral problem Christine Calculus 3 March 1st, 2013 08:56 AM
IM STUCK PLZ HELP... khalidd Algebra 1 September 9th, 2010 11:19 AM





Copyright © 2017 My Math Forum. All rights reserved.