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Inverse Trig Calc BC Integration

  • Thread starter RentonT
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  • #1
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An interesting AP Calculus BC problem I have not been able to solve.

Homework Statement


"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)


Homework Equations


sqrt(x)=sin(y)
integral(0,1/2,sqrt(x)/sqrt(1-x),dx)

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).

derivative(tan(y),dy)=ln(sec(y))+C
 
Last edited:

Answers and Replies

  • #2
Dick
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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.
 
  • #3
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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.
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
 
  • #4
vela
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You forgot about the dx. You need to write the integral in terms of dy.
 

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