Question

How do you integrate `sqrt(1- tan^2(x))` with respect to `x`?

Answer 1

Use WolframAlpha for complicated integrals, like this one.

Explanation:

I used WolframAlpha to do the integral.

`intsqrt(1 - tan^2(x))dx =`

`(cos(x) sqrt(1 - tan^2(x)) (sqrt(2) sin^(-1)(sqrt(2) sin(x)) - tan^(-1)((sin(x))/sqrt(cos(2 x)))))/sqrt(cos(2 x)) + C`

Answer 2

`sqrt2arctan((sqrt2tanx)/sqrt(1-tan^2x))-arcsin(tanx)+C`

Explanation:

`I=intsqrt(1-tan^2x)dx`

Let `sintheta=tanx` so `costhetad theta=sec^2xdx`.

`I=intsqrt(1-tan^2x)/sec^2x(sec^2xdx)`

`I=intsqrt(1-tan^2x)/(tan^2x+1)(sec^2xdx)`

`I=intsqrt(1-sin^2theta)/(1+sin^2theta)(costhetad theta)`

`I=intcos^2theta/(1+sin^2theta)d theta`

Write in terms of `tantheta` and `sectheta`:

`I=int(1/sec^2theta)/(1+tan^2theta/sec^2theta)d theta`

`I=int1/(sec^2theta+tan^2theta)d theta`

Let `sec^2theta=1+tan^2theta`:

`I=int1/(2tan^2theta+1)d theta`

Multiply by `sec^2theta/sec^2theta=sec^2theta/(tan^2theta+1)`:

`I=intsec^2theta/((tan^2theta+1)(2tan^2theta+1))d theta`

Let `u=tantheta` so `du=sec^2thetad theta`:

`I=int1/((u^2+1)(2u^2+1))du`

Omitting the process of partial fraction decomposition, this becomes:

`I=2int1/(2u^2+1)du-int1/(u^2+1)du`

Both of which are forms of arctangent integrals:

`I=sqrt2intsqrt2/((sqrt2u)^2+1)du-int1/(u^2+1)du`

`I=sqrt2arctan(sqrt2u)-arctan(u)`

Recall `u=tantheta`:

`I=sqrt2arctan(sqrt2tantheta)-arctan(tantheta)`

`I=sqrt2arctan(sqrt2tantheta)-theta`

And from `sintheta=tanx` we note that `theta=arcsin(tanx)` and `tantheta=sintheta/sqrt(1-sin^2theta)`:

`I=sqrt2arctan((sqrt2tanx)/sqrt(1-tan^2x))-arcsin(tanx)+C`