How to integrate this equation ∫dx/√(2ax - x² )?
- Thread starter Indranil
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what do you mean by 'Try completing the square on the radicand'. Could you explain, please?
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The radicand is the expression under the radical (in the integrand):
\(\displaystyle 2ax-x^2=-\left(x^2-2ax\right)=-\left(x^2-2ax+a^2\right)+a^2=a^2-(x-a)^2\)
And now we have:
\(\displaystyle \displaystyle \int\frac{1}{\sqrt{a^2-(x-a)^2}}\,dx\)
Can you proceed now with a trig. substitution?
No.
I understand well what you have done above but I don't know how to do trig. substitution. Could you help me with that, please?
Dr.Peterson
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Please show us what you have been able to do, and tell us what you have learned about integration. Have you ever done a trig substitution? Do you have a textbook that explains it? Have you read that?
HallsofIvy
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Hopefully you do know that, for any x, \(\displaystyle sin^2(x)+ cos^2(x)= 1\) so that \(\displaystyle cos^2(x)= 1- sin^2(x)\). That gives a "template" for converting square roots of something like "\(\displaystyle \sqrt{1- y^2}\)" to things that do not have the square root.
Here, the problem is to integrate \(\displaystyle \sqrt{a^2- (x- a)^2}\). The first step, prior to a "trig substitution", is to let y= x- a. Then dy= dx and the integrand becomes \(\displaystyle \sqrt{a^2- y^2}\). Now the trig substitution: let \(\displaystyle y= a sin(\theta)\). That way \(\displaystyle a^2- y^2= a^2- a^2 sin^2(\theta)= a^2(1- sin^2(\theta))= a^2 cos^2(\theta)\) and \(\displaystyle \sqrt{a^2- y^2}= \sqrt{a^2 cos^2(\theta)}= a cos(\theta)\). Of course, with \(\displaystyle y= a sin(\theta)\), \(\displaystyle dy= a cos(\theta) d\theta\).
Putting those together, \(\displaystyle \int \frac{1}{\sqrt{a^2- (x- a)^2}}dx= \int \frac{1}{\sqrt{a^2- y^2}}dy= \int \frac{1}{a cos(\theta} (cos(\theta) d\theta= a\int d\theta= a\theta+ C\).
Now, go back to x: since \(\displaystyle y= sin(\theta)\), \(\displaystyle \theta= arcsin(y)\) and the integral is \(\displaystyle a arcsin(y)+ C\). And since y= x- a, x= y+ a. We have \(\displaystyle a arcsin(y+ a)+ C\).
Here, the problem is to integrate \(\displaystyle \sqrt{a^2- (x- a)^2}\). The first step, prior to a "trig substitution", is to let y= x- a. Then dy= dx and the integrand becomes \(\displaystyle \sqrt{a^2- y^2}\). Now the trig substitution: let \(\displaystyle y= a sin(\theta)\). That way \(\displaystyle a^2- y^2= a^2- a^2 sin^2(\theta)= a^2(1- sin^2(\theta))= a^2 cos^2(\theta)\) and \(\displaystyle \sqrt{a^2- y^2}= \sqrt{a^2 cos^2(\theta)}= a cos(\theta)\). Of course, with \(\displaystyle y= a sin(\theta)\), \(\displaystyle dy= a cos(\theta) d\theta\).
Putting those together, \(\displaystyle \int \frac{1}{\sqrt{a^2- (x- a)^2}}dx= \int \frac{1}{\sqrt{a^2- y^2}}dy= \int \frac{1}{a cos(\theta} (cos(\theta) d\theta= a\int d\theta= a\theta+ C\).
Now, go back to x: since \(\displaystyle y= sin(\theta)\), \(\displaystyle \theta= arcsin(y)\) and the integral is \(\displaystyle a arcsin(y)+ C\). And since y= x- a, x= y+ a. We have \(\displaystyle a arcsin(y+ a)+ C\).
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I would let:
\(\displaystyle x-a=a\sin(\theta)\implies dx=a\cos(\theta)\,d\theta\)
Hence:
\(\displaystyle \displaystyle \int\frac{1}{\sqrt{a^2-(x-a)^2}}\,dx= \int\frac{a\cos(\theta)}{\sqrt{a^2-a^2\sin^2(\theta)}}\,d\theta=\int\,d\theta=\theta+C\)
Now, from our substitution we find:
\(\displaystyle \theta=\arcsin\left(\frac{x-a}{a}\right)\)
And so we have:
\(\displaystyle \displaystyle \int\frac{1}{\sqrt{2ax-x^2}}\,dx=\arcsin\left(\frac{x-a}{a}\right)+C\)
Note: I have assumed \(\displaystyle a\) is non-negative.
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I am sorry to say that I don't know how to do a trig substitution. I don't have any text book.
Dr.Peterson
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Then get one! A site like this is not meant to be where you first learn calculus, but where you can get help learning, as you work through some organized material.
One place to find textbook-like teaching is here. (In fact, it's the first hit I get when I search for "trig substitution", which I would hope you have done yourself.) If the material on that page assumes things you don't know yet, back up to previous chapters or even courses on the same site.
You've been shown two ways to do the substitution, so you don't need more from me on that; but you will benefit from seeing more background concerning why you make the choices you do.
Thank you for giving such link. Thanks a lot.
mmm4444bot
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There are many free math textbooks online. Whether for algebra, trigonometry, or calculus, try googling keywords like free online math textbooks for calculus.
For example, one hit is this textbook, which begins with simple trig substitutions in section 8.1 (page 156), and then goes on to show more involved methods in section 8.3 (bottom of page 162).
For example, one hit is this textbook, which begins with simple trig substitutions in section 8.1 (page 156), and then goes on to show more involved methods in section 8.3 (bottom of page 162).