`int(1/sqrt(z)+z^2/sqrt(z))dz=?`.

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 :

If e^(x)=(sqrt(1+z)-sqrt(1-z))/(sqrt(1+z)+sqrt(1-z)) and tan (y/2)=sqrt((1-z)/(1+z)) then the value of (dy)/(dx) at z=1 is equal to to

Prove that : tan^(-1) [ (sqrt(1+z) +sqrt(1-z))/(sqrt(1+z) -sqrt(1-z))] = pi/4 +1/2 cos^(-1) z

The value of Z is given by the following if z^(z sqrt(z))=(z sqrt(z))^(z)

Prove that int_0^(pi/2)(xdx)/(sinx)=int_0^1(sin^-1z)/(zsqrt(1-z^2))dz.

The value of z is given by the folowing if z^(z sqrt(z))=z sqrt(z)^(z)

int ((2z+1)/(5z+1))dz=

If |z_(1)| = sqrt(2), |z_(2)| = sqrt(3) and |z_(1) + z_(2)| = sqrt((5-2sqrt(3))) then arg ((z_(1))/(z_(2))) (not neccessarily principal)