Let I=intsqrttanxdxI=∫√tanxdx.
We subst. tanx=t^2", so that, "sec^2xdx=2tdttanx=t2, so that, sec2xdx=2tdt, or,
dx=(2tdt)/sec^2x=(2tdt)/(1+tan^2x)=(2tdt)/(1+t^4).dx=2tdtsec2x=2tdt1+tan2x=2tdt1+t4. Hence,
I=int{t*(2t)/(1+t^4)}dt=int(2t^2)/(1+t^4)dt=int2/(t^2+1/t^2)dtI=∫{t⋅2t1+t4}dt=∫2t21+t4dt=∫2t2+1t2dt
=int((1+1/t^2)+(1-1/t^2))/(t^2+1/t^2)dt=∫(1+1t2)+(1−1t2)t2+1t2dt
=int(1+1/t^2)/(t^2+1/t^2)dt+int(1-1/t^2)/(t^2+1/t^2)dt=I_1+I_2, say=∫1+1t2t2+1t2dt+∫1−1t2t2+1t2dt=I1+I2,say,
where, I_1=int(1+1/t^2)/(t^2+1/t^2)dt.I1=∫1+1t2t2+1t2dt.
Subst. t-1/t=u rArr (1+1/t^2)dt=du, &, t^2+1/t^2=u^2+2.t−1t=u⇒(1+1t2)dt=du,&,t2+1t2=u2+2.
:. I_1=int1/(u^2+(sqrt2)^2)du=1/sqrt2arc tan(u/sqrt2).
For, I_2=int(1-1/t^2)/(t^2+1/t^2)dt,
we take, (t+1/t)=v rArr (1-1/t^2)dt=dv, &, t^2+1/t^2=v^2-2.
:. I_2=int1/(v^2-(sqrt2)^2)dv=1/(2sqrt2)*ln|(v-sqrt2)/(v+sqrt2)|.
Thus, I_1=1/sqrt2arc tan((t-1/t)/sqrt2)=1/sqrt2arc tan{(t^2-1)/(sqrt2t)}
=1/sqrt2arc tan{(tanx-1)/sqrt(2tanx)}, and,
I_2=1/(2sqrt2)ln|(t+1/t-sqrt2)/(t+1/t+sqrt2)|
=1/(2sqrt2)ln|(t^2-sqrt2t+1)/(t^2+sqrt2t+1)|
=1/(2sqrt2)ln|(tanx-sqrt(2tanx)+1)/(tanx+sqrt(2tanx)+1)|.
Finally, I=I_1+I_2
=1/sqrt2arc tan{(tanx-1)/(sqrt(2tanx))}
+1/(2sqrt2)ln|(tanx-sqrt(2tanx)+1)/(tanx+sqrt(2tanx)+1)|+C.
Enjoy Maths.!
