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Solving an integral using an special substitution

  • Thread starter Telemachus
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  • #1
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Homework Statement


Well, the exercise asks me to solve the next integral using an adequate substitution.
[tex]\displaystyle\int_{}^{}\sqrt[ ]{4-x^2}dx[/tex]

The Attempt at a Solution


[tex]\displaystyle\int_{}^{}\sqrt[ ]{4-x^2}dx[/tex]

What I did was:

[tex]x=2\sin\theta[/tex]

[tex]dx=2\cos\theta d\theta[/tex]

So, then I get
[tex]\displaystyle\int_{}^{}\sqrt[ ]{4-x^2}dx=4\displaystyle\int_{}^{}\cos^2\theta d\theta=cos\theta\sin\theta+\displaystyle\int_{}^{}\sin^2\theta d\theta=cos\theta\sin\theta+\displaystyle\int_{}^{}(1-\cos^2\theta) d\theta=[/tex]
[tex]=cos\theta\sin\theta+\displaystyle\int_{}^{}d\theta-\displaystyle\int_{}^{}\cos^2\theta[/tex]

[tex]\Rightarrow{5\displaystyle\int_{}^{}\cos^2\theta d\theta=cos\theta\sin\theta+\theta}\Rightarrow{\displaystyle\int_{}^{}\cos^2\theta d\theta=\displaystyle\frac{1}{5}cos\theta\sin\theta+\displaystyle\frac{\theta}{5}}[/tex]

[tex]\theta=\arcsin(\displaystyle\frac{x}{2})[/tex]
[tex]\sin\theta=(\displaystyle\frac{x}{2})[/tex]

[tex]\cos\theta=\sqrt[ ]{1-\sin^2\theta}[/tex]

[tex]\cos\theta=\sqrt[ ]{1-\displaystyle\frac{x^2}{4}}[/tex]

[tex]\displaystyle\int_{}^{}\sqrt[ ]{4-x^2}=\displaystyle\frac{x}{10}\sqrt[ ]{1-\displaystyle\frac{x^2}{4}}+\displaystyle\frac{\arcsin(\displaystyle\frac{x}{2})}{5}[/tex]

I think I've made a mistake, but I don't know in which step. Perhaps when I've get back to the original variables, or when I've solved by parts the integral of the square of the cosine.

Bye there, and thanks for posting.
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


Well, the exercise asks me to solve the next integral using an adequate substitution.
[tex]\displaystyle\int_{}^{}\sqrt[ ]{4-x^2}dx[/tex]

The Attempt at a Solution


[tex]\displaystyle\int_{}^{}\sqrt[ ]{4-x^2}dx[/tex]

What I did was:

[tex]x=2\sin\theta[/tex]

[tex]dx=2\cos\theta d\theta[/tex]

So, then I get
[tex]\displaystyle\int_{}^{}\sqrt[ ]{4-x^2}dx=4\displaystyle\int_{}^{}\cos^2\theta d\theta[/tex]

[
Right there, try using the identity

[tex]\cos^2\theta = \frac{1+\cos(2\theta)}{2}[/tex]
 
  • #3
832
30
Thank you LCKurtz.
 
  • #4
832
30
Can somebody tell me if this is right?

I must solve:
[tex]\displaystyle\int_{}^{}\displaystyle\frac{dx}{\sqrt[ ]{9+x^2}}[/tex]

And I used the substitutions:
[tex]x=3sinht[/tex]
[tex]dx=3coshtdt[/tex]

And the identity:

[tex]cosh^2t-sinh^2t=1\Rightarrow{\cosht=\sqrt[ ]{1+sinh^2t}}[/tex]

[tex]\displaystyle\int_{}^{}\displaystyle\frac{dx}{\sqrt[ ]{9+x^2}}=\displaystyle\int_{}^{}\displaystyle\frac{3coshtdt}{\sqrt[ ]{3^2+(3sinht)^2}}=\displaystyle\int_{}^{}\displaystyle\frac{3coshtdt}{\sqrt[ ]{3^2(1+sinh^2t)}}=\displaystyle\int_{}^{}\displaystyle\frac{coshtdt}{cosht}=\displaystyle\int_{}^{}dt=t+C=argsh(\displaystyle\frac{x}{3})+C[/tex]
 
Last edited:
  • #5
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I don't see anything obviously wrong, but I would have used a different substitution, tan(theta) = x/3.
 
  • #6
832
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Thanks Mark.

What about this one? I don't know how to proceed now:

[tex]\displaystyle\int_{}^{}\displaystyle\frac{dt}{\sin(t)}[/tex]

I've solved this using the table, with definition of cosecant: [tex]\displaystyle\int_{}^{}\displaystyle\frac{dt}{\sin(t)}=\ln|csc(t)-cot(t)|+C[/tex]

This was the result of solving: [tex]\displaystyle\int_{}^{}\displaystyle\frac{dx}{x\sqrt[ ]{1-x^2}}=\ln|csc(arcsin(x))-cot(arcsin(x))|+C[/tex]

But now...
[tex]\displaystyle\int_{}^{}\displaystyle\frac{dx}{x^2\sqrt[ ]{4-x^2}}[/tex]

[tex]x=2\sin(t)[/tex]
[tex]dx=2\cos(t)dt[/tex]

[tex]\displaystyle\int_{}^{}\displaystyle\frac{dx}{x^2\sqrt[ ]{4-x^2}}=\displaystyle\int_{}^{}\displaystyle\frac{2\cos(t)dt}{4\sin^2(t)\sqrt[ ]{2^2-(2\sin(t)^2}}=\displaystyle\frac{1}{4}\displaystyle\int_{}^{}\displaystyle\frac{dt}{\sin^2(t)}}=\displaystyle\frac{1}{4}\displaystyle\int_{}^{}csc^2(t)dt[/tex]

I thought of solving by parts, but...

[tex]u=csc(t)[/tex]
[tex]du=-csc(t)cot(t)dt[/tex]

[tex]dv=csc(t)dt[/tex]
[tex]v=\ln|csc(t)-cot(t)|[/tex]
 
  • #7
LCKurtz
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[tex]\displaystyle\int_{}^{}\displaystyle\frac{dx}{x^2\sqrt[ ]{4-x^2}}=\displaystyle\int_{}^{}\displaystyle\frac{2\cos(t)dt}{4\sin^2(t)\sqrt[ ]{2^2-(2\sin(t)^2}}=\displaystyle\frac{1}{4}\displaystyle\int_{}^{}\displaystyle\frac{dt}{\sin^2(t)}}=\displaystyle\frac{1}{4}\displaystyle\int_{}^{}csc^2(t)dt[/tex]

I thought of solving by parts, but...
Hint: What is the derivative of cot(t)?

Also Forum procedures suggest that you post new questions in a new thread.
 
  • #8
832
30
Thanks LCKurtz. Srry for posting here again, is just that is the same kind of problem, I didn't wanted to flood with topics of mine.

I've arrived to this solution:
[tex]\displaystyle\frac{1}{4}\displaystyle\int_{}^{}csc^2(t )dt=-\displaystyle\frac{1}{4}cot(arcsin(\displaystyle\frac{x}{2}))[/tex]
 
  • #9
Hurkyl
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Right there, try using the identity

[tex]\cos^2\theta = \frac{1+\cos(2\theta)}{2}[/tex]
For the record, the method the opening poster used (integration by parts) is perfectly valid, and IMHO fairly clever. His problem there is that he forgot the 4 outside of the inetgral.
 
  • #10
832
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Thanks, I've solved it both ways :)

Using that substitution drives faster to the solution.
 
  • #11
hunt_mat
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Another way (and I believe that this is the classical way of looking at it)
[tex]
I=\int\frac{dx}{\sin x}=\int\frac{\sin xdx}{\sin^{2}x}=\int\frac{\sin xdx}{1-\cos^{2}x}
[/tex]
Then split using partial fractions to obtain:
[tex]
I=\frac{1}{2}\int\frac{\sin x}{1-\cos x}dx+\frac{1}{2}\int\frac{\sin x}{1+\cos x}dx
[/tex]
The integral may be easily computed to arrive at:
[tex]
I=-\frac{1}{2}\log (1-\cos x)+\frac{1}{2}\log (1+\cos x)
[/tex]
Now with a bit of algebra which includes a bit of trig identities, we arrive at the answer:
[tex]
I=\log |\cot x+cosec x|
[/tex]
 
Last edited:
  • #12
832
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Thank you hunt_mat. I always enjoy seeing the many perspectives that surges from facing a problem in this forum.
 
  • #13
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In case anyone has lost track, the work below refers to the integral in post #6, not the one in the OP.
Another way (and I believe that this is the classical way of looking at it)
[tex]
I=\int\frac{dx}{\sin x}=\int\frac{\sin xdx}{\sin^{2}x}=\int\frac{\sin xdx}{1-\cos^{2}x}
[/tex]
Then split using partial fractions to obtain:
[tex]
I=\frac{1}{2}\int\frac{\sin x}{1-\cos x}dx+\frac{1}{2}\int\frac{\sin x}{1+\cos x}dx
[/tex]
The integral may be easily computed to arrive at:
[tex]
I=-\frac{1}{2}\log (1-\cos x)+\frac{1}{2}\log (1+\cos x)
[/tex]
Now with a bit of algebra which includes a bit of trig identities, we arrive at the answer:
[tex]
I=\log |\cot x+cosec x|
[/tex]
A different approach is the following:
[tex]\int\frac{dx}{\sin x} = \int csc~x~dx = \int csc~x\frac{csc~x + cot~x}{csc~x + cot~x}dx[/tex]
[tex]= \int \frac{csc^2~x + csc~x cot~x}{csc~x + cot~x}dx[/tex]

Now you can use an ordinary substitution, u = csc(x) + cot(x), du = -csc2(x) - csc(x)cot(x)dx, to get -ln|csc(x) + cot(x)| + C.
 
  • #14
hunt_mat
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Mine was easier. :eek:)
 
  • #15
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I don't think so. Mine involved multiplying by 1 and an ordinary substitution. Yours requires partial fraction decomposition.
 
  • #16
Hurkyl
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It's even easier to just substitute -ln|csc(x)+cot(x)|+C for the integral. :tongue:
 
  • #17
hunt_mat
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The multiplication by 1 bit was inspired, I wouldn't have thought about doing that. All the inspired bit that I did was again multiply by 1 but this time it was much more intuitive. Partial fraction are something everyone should know, it's a simple technique that is easily applied.
 

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