`int_0^(pi/3)(dz)/(sqrt(e^z))`

int_(0)^(pi//2)(dx)/(1+e^(sqrt(2)cos(x+(pi)/(4)))) is equal to

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

int_0^(pi) sqrt x dx =

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

lim_(hto0)(1)/(h)int_(x)^(x+h)(dz)/(z+sqrt(z^(2)+1)) is equal to -

Prove that pi/6 < int_0^1(dx)/(sqrt(4-x^2-x^3)) < pi/(4sqrt(2))

Prove that pi/6

inte^(2z)sqrt(e^(4z)+6)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 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 :