1.

If ∫√1+2tanx(tanx+secx)dx=ln|f(x)|+C, then the value of f(0) is equal to (0≤x<π2)(where C is constant of integration )

Answer» If 1+2tanx(tanx+secx)dx=ln|f(x)|+C, then the value of f(0) is equal to (0x<π2)

(where C is constant of integration )




Discussion

No Comment Found