`int1/(7+5cosx)\ dx=` `1/(sqrt(6))tan^(-1)(1/(sqrt(6))tanx/2)+C` (b) `1/(sqrt(3))tan^(-1)(1/(sqrt(3))tanx/2)+C` (c) `1/4tan^(-1)(tanx/2)+C` (d) `1/7tan^(-1)(tanx/2)+C`

int(1)/(7+5cos x)dx=(1)/(sqrt(6))tan^(-1)((1)/(sqrt(6))(tan x)/(2))+C(b)(1)/(sqrt(3))tan^(-1)((1)/(sqrt(3))(tan x)/(2))+C(c)(1)/(4)tan^(-1)((tan x)/(2))+C(d)(1)/(7)tan^(-1)((tan x)/(2))+C

int(1)/(1+cos^(2)x)dx=(1)/(sqrt(2))tan^(-1)((tan x)/(sqrt(2)))+c

tan^(2)x -(1+sqrt(3))tanx + sqrt(3)=0

The value of tan^(-1)((sqrt(3))/(2))+tan^(-1)((1)/(sqrt(3))) is equal to a) tan^(-1)((5)/(sqrt(3))) b) tan^(-1)((2)/(sqrt(3))) c) tan^(-1)((1)/(2)) d) tan^(-1)((1)/(3sqrt(3)))

int(dx)/(4sin^(2)x+3cos^(2)x) is equal to a) (sqrt(3))/(4)tan^(-1)""((2tanx)/(sqrt(3)))+C b) (1)/(2sqrt(3))tan^(-1)""((tanx)/(sqrt(3)))+C c) (2)/(sqrt(3))tan^(-1)""((2tanx)/(sqrt(3)))+C d) (1)/(2sqrt(3))tan^(-1)""((2tanx)/(sqrt(3)))+C

int(sinx)/(3+4cos^2x)\ dx a. log(3+4cos^2x)+C (b) 1/(2sqrt(3))tan^(-1)((cosx)/(sqrt(3)))+C (c) -1/(2sqrt(3))tan^(-1)((2"\ cos"x)/(sqrt(3)))+C (d) 1/(2sqrt(3))tan^(-1)((2"\ cos"x)/(sqrt(3)))+C

(d)/(dx)[tan^(-1)((2+3tanx)/(3-2tanx))]=

Show that tan^(-1)(1/sqrt3tanx/2)=1/2 cos^(-1)(1+2cosx)/(2+cosx) .

int(dx)/(3+sin2x)=(1)/(sqrt(k))Tan^(-1)((3Tan x+1)/(sqrt(8)))+C where k is

int (dx)/((x + 1) sqrt(x)) is equal to a) tan^(-1) sqrt(x) + C b) 2 tan^(-1) x + C c) 2 tan^(-1) (sqrt(x)) + C d) tan^(-1) (x^((3)/(2))) + C