Question

How do you integrate `sqrttanx`?

Answer 1

`1/sqrt2arc tan{(tanx-1)/(sqrt(2tanx))}`
`+1/(2sqrt2)ln|(tanx-sqrt(2tanx)+1)/(tanx+sqrt(2tanx)+1)|+C`.

Explanation:

Let `I=intsqrttanxdx`.

We subst. `tanx=t^2", so that, "sec^2xdx=2tdt`, or,

`dx=(2tdt)/sec^2x=(2tdt)/(1+tan^2x)=(2tdt)/(1+t^4).` Hence,

`I=int{t*(2t)/(1+t^4)}dt=int(2t^2)/(1+t^4)dt=int2/(t^2+1/t^2)dt`

`=int((1+1/t^2)+(1-1/t^2))/(t^2+1/t^2)dt`

`=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`,

where, `I_1=int(1+1/t^2)/(t^2+1/t^2)dt.`

Subst. `t-1/t=u rArr (1+1/t^2)dt=du, &, t^2+1/t^2=u^2+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.!

Answer 2

`I=int sqrttanxdx`

`=1/2int (2sqrttanx)dx`

`=1/2[int (sqrttanx+sqrtcotx)dx+ int(sqrttanx-sqrtcotx)dx]`

Let
`I_1=int (sqrttanx+sqrtcotx)dx`

`=sqrt2int (sinx+cosx)/sqrt(2sinxcosx)dx`

`=sqrt2int (sinx+cosx)/sqrt(1-(sinx-cosx)^2)dx`

If `sinx -cosx =u " then " du =(sinx+cosx)dx`

So

`I_1=sqrt2int1/sqrt(1-u^2) du=sqrt2sin^-1u`
`=sqrt2sin^-1(sinx-cosx)`

Similarly
`I_2=int (sqrttanx-sqrtcotx)dx`

`=sqrt2int (sinx-cosx)/sqrt(2sinxcosx)dx`

`=-sqrt2int (cosx-sinx)/sqrt((sinx+cosx)^2-1)dx`

If `sinx +cosx =v" then " dv =(cosx-sinx)dx`

So

`I_2=-sqrt2int1/sqrt(v^2-1) dv=-sqrt2lnabs(v+sqrt(v^2-1))`
`=-sqrt2(lnabs((sinx+cosx)+sqrt(2sinxcosx)))`

`I=1/2(I_1+I_2)`

`=1/2[sqrt2sin^-1(sinx-cosx)-sqrt2lnabs((sinx+cosx)+sqrt(2sinxcosx))]+C`

`=1/sqrt2[sin^-1(sinx-cosx)-lnabs((sinx+cosx)+sqrt(2sinxcosx))]+C`