Let, I=int(sqrtcotx-sqrttanx)/(1+3sin2x)dx.I=∫√cotx−√tanx1+3sin2xdx.
:. I=int(1-tanx)/(sqrttanx(1+3sin2x))dx.
Using, sin2x=(2tanx)/(1+tan^2x), we have,
1+3sin2x=1+(6tanx)/(1+tan^2x)=(tan^2x+6tanx+1)/(1+tan^2x), so,
I=int{(1-tanx)(1+tan^2x)}/{sqrttanx(tan^2x+6tanx+1)}dx, or,
I=int{(1-tanx)(sec^2xdx)}/{sqrttanx(tan^2x+6tanx+1)}.
Now, we use the substn. tanx=t^2rArr sec^2xdx=2tdt.
:. I=int{(1-t^2)(2tdt)}/{t(t^4+6t^2+1)}=2int(1-t^2)/(t^4+6t^2+1)dt.
:. I=2int{t^2(1/t^2-1)}/{t^2(t^2+6+1/t^2)dt
=-2int(1-1/t^2)/{(t+1/t)^2+4}dt.
Finally, sub.ing, t+1/t=u rArr (1-1/t^2)dt=du.
I=-2int1/(u^2+4)du.
=-2(1/2)arc tan(u/2)=-arc tan(u/2).=-arc tan{1/2(t+1/t)}
:. I=-arc tan {1/2(sqrttanx+sqrtcotx)}+C.
Enjoy Maths.!
