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An interesting AP Calculus BC problem I have not been able to solve.

"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)

sqrt(x)=sin(y)

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

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

## 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

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