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Help with trigonometric substitutions

  1. Aug 21, 2008 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    √a^2-x^2 , x=a sin (theta) , -pi/2 less than or equal theta less than or equal pi/2 Identity 1-sin^2 theta= cos^2 theta

    table of trigonometric substitutions

    3. The attempt at a solution
    x=5 sin theta
    dx= 5 cos theta dtheta
    theta=arcsin (x/5)
    √25-x^2=5 cos theta
    ∫5 cos theta / 25 sin^2 theta * (5cos theta)=1/25 ∫sin^2 theta d theta

    =1/25 - cos^2 theta+c
    = -1/25 cos^2 (arcsin(x/5) +c

    but the answer in the back of the book is -√25-(x^2)/(25x) +c

    what did i do wrong
  2. jcsd
  3. Aug 21, 2008 #2
    anyone know what i did wrong
  4. Aug 22, 2008 #3


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    Homework Helper

    Is the integral

    [tex]\int \frac{dx}{x^2 \cdot \surd(25-x^2)} [/tex]?

    How did sin^2 suddenly get into the numerator? I believe the trig-substituted integral at this point will be

    [tex]\frac{1}{25} \int \frac{d\theta}{sin^2 \theta}[/tex]

    Do we have an antiderivative handy for [tex]csc^2 \theta[/tex]?

    BTW, please don't "bump" your own threads: it makes the reply count go up, so it looks like someone has started helping you when they actually haven't yet. Please be patient and someone will get to you. It's going to be slow right now because most universities are still on "interim" until at least next week (and many don't start up again until well into September)...
    Last edited: Aug 22, 2008
  5. Aug 22, 2008 #4


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    Science Advisor

    First question: is this
    [tex]\int \frac{dx}{x^2\sqrt{25- x^2}}[/tex]
    [tex]\int \frac{1}{x^2} \sqrt{25- x^2}dx[/tex]?

    If the problem is
    [tex]\int \frac{dx}{x^2\sqrt{25- x^2}}[/tex]
    then the substitution x= 5 sin(t) gives dx= 5 cos(t)dt, [itex]\sqrt{25- x^2}= 5cos(t)[/itex] and x2= 25 sin2(t) so the integral becomes
    [tex]\int \frac{5 cos(t)dt}{(25 sin^2(t))(5cos(t))}= \frac{1}{25}\int \frac{dt}{sin^2(t)}= \frac{1}{25}\int csc^2(t) dt[/tex]

    If the problem is
    [tex]int \frac{1}{x^2}\sqrt{25- x^2}dx[/tex]
    then the substitution x= 5 sin(t) gives
    [tex]\int \frac{25 cos^2(t) dt}{25 sin^2(t)}= \int cot^2(t) dt[/itex]
  6. Aug 22, 2008 #5
    It's the first one
  7. Aug 23, 2008 #6


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    Homework Helper

    Then the replies in post #3 and the first integration in post #4 are appropriate. Find the general antiderivative of [tex]
    csc^2 \theta
    [/tex], then back-substitute to get the expression in terms of x.
  8. Aug 23, 2008 #7
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