If `y=sqrt((t-alpha)/(beta-t)) and x=sqrt((t-alpha)(beta-t))` then find `(dy)/(dx).`

int _(alpha)^(beta) sqrt((x-alpha)/(beta -x)) dx is equal to

int _(alpha)^(beta) sqrt((x-alpha)/(beta -x)) dx is equal to

If x=(sin^(3)t)/(sqrt(cos2t)) and y=(cos^(3)t)/(sqrt(cos2t)) , then find (dy)/(dx) .

If x=sqrt( a^(sin^(-1)t)) and y=sqrt( a^(cos^(-1)t)) find dy/dx

If e^x=(sqrt(1+t)-sqrt(1-t))/(sqrt(1+t)+sqrt(1-t)) and tan (y/2)=sqrt((1-t)/(1+t)) then (dy)/dx at t=1/2 is

Computing area with parametrically represented boundaries : If the boundary of a figure is represented by parametric equation, i.e., x=x(t), y=y(t), then the area of the figure is evaluated by one of the three formulas : S=-int_(alpha)^(beta)y(t)x'(t)dt, S=int_(alpha)^(beta)x(t)y'(t)dt, S=(1)/(2)int_(alpha)^(beta)(xy'-yx')dt, Where alpha and beta are the values of the parameter t corresponding respectively to the beginning and the end of the traversal of the curve corresponding to increasing t. The area of the loop described as x=(t)/(3)(6-t),y=(t^(2))/(8)(6-t) is