My Math Forum integrals
 User Name Remember Me? Password

 Calculus Calculus Math Forum

 January 16th, 2013, 09:38 AM #1 Senior Member   Joined: Oct 2012 Posts: 460 Thanks: 0 integrals [color=#0000FF]Dear All! Please, does anyone have an idea how to solve: $\int\sqrt{\frac{arcsinx}{1-x^2}}dx$ the result is ment to be: $\frac{2}{3}arcsinx^{\frac{3}{2}}$ Many thanks![/color]
 January 16th, 2013, 10:26 AM #2 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: integrals Use the substitution t = arcsin(x) ,
 January 16th, 2013, 09:25 PM #3 Newbie   Joined: Jan 2013 Posts: 13 Thanks: 0 Re: integrals Interesting problem. It's just a u-sub as zaida mentioned. Split the $\sqrt{u/1-x^2}$ into $\sqrt{u}/\sqrt{1-x^2}$ if you don't see the cancelling. Jc
 January 19th, 2013, 07:57 AM #4 Senior Member   Joined: Oct 2012 Posts: 460 Thanks: 0 Re: integrals [color=#0000FF]Thank you! now I have the integral of a rational function! $\int\frac{\sqrt{u}}{\sqrt{1-x^2}}$ what now? http://integral-table.com/integral-tabl ... 0000000000 nth frm this goes well with...??? [/color]
January 20th, 2013, 02:54 PM   #5
Math Team

Joined: Sep 2007

Posts: 2,409
Thanks: 6

Re: integrals

Quote:
Originally Posted by ungeheuer
[color=#0000FF][b]Thank you!
now I have the integral of a rational function!

$\int\frac{\sqrt{u}}{\sqrt{1-x^2}}[/color]$[color=#000000]
Nonsense! For one thing this is NOT a rational function because it involves square roots. Second, you do not have a "dx" or "du", necessary for an integral. Third, you have both "x" and "u" in the integral- you must have one or the other (depending upon whether you have "dx" or "du") not both.

What you had originally was $\int\sqrt{\frac{arcsin(x)}{1- x^2}}dx$ and the substitution $u= arcsin(x)$ was suggested. But you can't just replace the one instance of arcsin(x) by u and leave the rest of the integral alone. You will have to replace "dx" with something involving only "u" and "du". To get "dx" and "du", of course, you will need to take the derivative. What is the derivative of arcsin(x)?[/color]

Quote:
 [color= #0000FF] what now? http://integral-table.com/integral-tabl ... 0000000000 nth frm this goes well with...??? [/color]

 January 21st, 2013, 06:16 AM #6 Senior Member   Joined: Oct 2012 Posts: 460 Thanks: 0 Re: integrals [color=#0000FF]Thank U, HallsofIvvy! (Well, be honest and admit it is not that easy!) $\frac{d}{dx}arcsin(x)=\frac{1}{\sqrt{}1-x^2}$[/color]
 January 21st, 2013, 06:28 AM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: integrals $y\,=\,\arcsin(x)$ $sin(y)\,=\,x$ $y'\cos(y)\,=\,1$ $y'\,=\,\frac{1}{\sqrt{1\,-\,x^2}}$
 January 21st, 2013, 06:37 AM #8 Senior Member   Joined: Oct 2012 Posts: 460 Thanks: 0 Re: integrals $y'cos(y)=1$[color=#0000FF] I dont get this... we dont know what is cosy? how do we know that it is 1?[/color]
 January 23rd, 2013, 10:14 AM #9 Senior Member   Joined: Oct 2012 Posts: 460 Thanks: 0 Re: integrals [color=#0000FF]Hello! Now I have: $\int \sqrt{\frac{arcsin(x)}{1-x^2}}= \int \sqrt{\frac{u}{1-x^2}}\frac{du}{1-x^2}$ what shoud I do now? many thanks! Not finding the idea at: http://integral-table.com/ [/color]
 January 23rd, 2013, 01:20 PM #10 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: integrals I believe you have done this same thing repeatedly and been told, repeatedly, that you cannot do it! You cannot replace part of the function so that it is in terms of "u" while leaving the rest in terms of "x". That was the whole point in suggesting the substitution u= arcsin(x). Differentiating both sides, $du= \frac{dx}{\sqrt{1- x^2}}$. You have put that in backwards, writing it as if it were $dx= \frac{du}{\sqrt{1- x^2}}$. Instead write the integral as $\int \sqrt{arcsin(x)}\frac{dx}{\sqrt{1- x^2}}= \int \sqrt{u}du$.

 Tags integrals

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post ungeheuer Calculus 10 July 21st, 2013 03:30 AM veronicak5678 Real Analysis 1 May 1st, 2012 12:25 PM damar10 Calculus 1 April 25th, 2012 08:34 AM johnnyboy20 Calculus 2 May 24th, 2011 04:36 PM hector manuel Real Analysis 0 May 4th, 2009 11:16 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top