# Inverse Trig Calc BC Integration

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

## Homework Statement

"If the substitution $$\sqrt(x)=sin(y)$$ is made in the integrand of $$\int\frac{\sqrt(x)*dx}{\sqrt(1-x)}$$ , 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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Dick
Homework Helper
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.

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

vela
Staff Emeritus