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Inverse Trig Calc BC Integration

  1. Mar 4, 2010 #1
    An interesting AP Calculus BC problem I have not been able to solve.

    1. The problem statement, all variables and given/known data
    "If the substitution [tex]\sqrt(x)=sin(y)[/tex] is made in the integrand of [tex]\int\frac{\sqrt(x)*dx}{\sqrt(1-x)}[/tex] , the resulting integral is ... [5 choices are given]

    (A) integral(0,1/2,(sin(y))^2,dy)
    (B) 2*integral(0,1/2,(sin(y))^2/cos(y),dy)
    (C) 2*integral(0,pi/4,(sin(y))^2,dy)
    (D) integral(0,pi/4,(sin(y))^2,dy)
    (E) 2*integral(0,pi/6,(sin(y))^2/cos(y),dy)

    2. Relevant equations

    3. The attempt at a solution
    Constructing a triangle yields a few values where 1 is the hypotenuse, sqrt(x) is opposite angle y and sqrt(1-x) is adjacent to y.

    The attempt was trigonometric substitution in the form of integral(sec(y)*sin(y)) which yields tangent y. Unfortunately, I do not know how to change the limits of integration and none of the answers are similar to the integral of tan(y).

    Last edited: Mar 4, 2010
  2. jcsd
  3. Mar 4, 2010 #2


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    Your substitution is x=sin(y)^2. One of those options is correct. How did you get sec(y)*sin(y)? That's not right.
  4. Mar 5, 2010 #3
    I constructed a triangle using the given information. The sin(y) as in the opposite leg over the hypotenuse is equal to sqrt(x)/1
  5. Mar 5, 2010 #4


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    You forgot about the dx. You need to write the integral in terms of dy.
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