Solve `l=int_(cos^(4)t)^(-sin^(4)t)(sqrt(f(z))dz)/(sqrt(f(cos 2 t -z))+sqrt(f(z)))`

x=sqrt(sin 2t),y=sqrt(cos 2 t)

f(x)=int_(0)^(x^(2))((sin^(-1)sqrt(t))^(2))/(sqrt(t))dt

int(dz)/(z sqrt(z^(2)-1))

If x=int_(0)^(t^(2))e^(sqrt(z)){(2tan sqrt(z)+1-tan^(2)sqrt(z))/(2sqrt(z)sec^(2)sqrt(z))}dz and y=int_(0)^(t^(2))e^(sqrt(z)){(1-tan^(2)sqrt(z)-2tan sqrt(z))/(2sqrt(z)sec^(2)sqrt(z))}dz : Then the inclination of the tangent to the curve at t=(pi)/(4) is :

If x=int_(0)^(t^(2))e^(sqrt(z)){(2tan sqrt(z)+1-tan^(2)sqrt(z))/(2sqrt(z)sec^(2)sqrt(z))}dz and x=int_(0)^(t^(2))e^(sqrt(z)){(1-tan^(2)sqrt(z)-2tan sqrt(z))/(2sqrt(z)sec^(2)sqrt(z))}dz : Then the inclination of the tangent to the curve at t=(pi)/(4) is :

x = (sin^3t)/(sqrt(cos 2t)), y = (cos^3 t)/(sqrt(cos 2t)) .

int_(0)^(sin^(2)x) sin^(-1) sqrt(t) dt + int_(0)^(cos^(2)x) cos^(-1) sqrt(t) dt=

If int_(x^(2))^(x^(4)) sin sqrt(t) dt, f'(x) equals

If int_(x^(2))^(x^(4)) sin sqrt(t) dt, f'(x) equals