Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Solving an integral using an special substitution

  1. Jun 27, 2010 #1
    1. The problem statement, all variables and given/known data
    Well, the exercise asks me to solve the next integral using an adequate substitution.
    [tex]\displaystyle\int_{}^{}\sqrt[ ]{4-x^2}dx[/tex]

    3. 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.
     
  2. jcsd
  3. Jun 27, 2010 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Right there, try using the identity

    [tex]\cos^2\theta = \frac{1+\cos(2\theta)}{2}[/tex]
     
  4. Jun 28, 2010 #3
    Thank you LCKurtz.
     
  5. Jun 28, 2010 #4
    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: Jun 28, 2010
  6. Jun 28, 2010 #5

    Mark44

    Staff: Mentor

    I don't see anything obviously wrong, but I would have used a different substitution, tan(theta) = x/3.
     
  7. Jun 28, 2010 #6
    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]
     
  8. Jun 28, 2010 #7

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hint: What is the derivative of cot(t)?

    Also Forum procedures suggest that you post new questions in a new thread.
     
  9. Jun 28, 2010 #8
    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]
     
  10. Jun 28, 2010 #9

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
  11. Jun 28, 2010 #10
    Thanks, I've solved it both ways :)

    Using that substitution drives faster to the solution.
     
  12. Jun 30, 2010 #11

    hunt_mat

    User Avatar
    Homework Helper

    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: Jun 30, 2010
  13. Jun 30, 2010 #12
    Thank you hunt_mat. I always enjoy seeing the many perspectives that surges from facing a problem in this forum.
     
  14. Jun 30, 2010 #13

    Mark44

    Staff: Mentor

    In case anyone has lost track, the work below refers to the integral in post #6, not the one in the OP.
    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.
     
  15. Jun 30, 2010 #14

    hunt_mat

    User Avatar
    Homework Helper

    Mine was easier. :eek:)
     
  16. Jun 30, 2010 #15

    Mark44

    Staff: Mentor

    I don't think so. Mine involved multiplying by 1 and an ordinary substitution. Yours requires partial fraction decomposition.
     
  17. Jun 30, 2010 #16

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It's even easier to just substitute -ln|csc(x)+cot(x)|+C for the integral. :tongue:
     
  18. Jul 1, 2010 #17

    hunt_mat

    User Avatar
    Homework Helper

    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook