Question

How do you find the integral of `1 / (1 + sin^2 x)`?

Answer 1

`arctan(sqrt2tanx)/sqrt2+C`

Explanation:

`I=int1/(1+sin^2x)dx`

Notice that dividing by `cos^2(x)` will get us working with `tanx` and `secx` functions, which are derivatives of themselves:

`I=int(1/cos^2x)/((1+sin^2x)/cos^2x)dx=intsec^2x/(sec^2x+tan^2x)dx`

Recall that `sec^2x=1+tan^2x`:

`I=intsec^2x/((1+tan^2x)+tan^2x)dx=intsec^2x/(1+2tan^2x)dx`

We will use the substitution `tanx=1/sqrt2tantheta`. This implies that `sec^2xdx=1/sqrt2sec^2thetad theta`. Substituting:

`I=int(1/sqrt2sec^2theta)/(1+2(1/sqrt2tantheta)^2)d theta=1/sqrt2int(sec^2theta)/(1+2(1/2tan^2theta))d theta`

`I=1/sqrt2int(sec^2theta)/(1+tan^2theta)d theta`

Using the same trigonometric identity:

`I=1/sqrt2intsec^2theta/sec^2thetad theta=1/sqrt2intd theta=1/sqrt2theta+C`

From `tanx=1/sqrt2tantheta` we see that `sqrt2tanx=tantheta` and `theta=arctan(sqrt2tanx)`. Thus:

`I=arctan(sqrt2tanx)/sqrt2+C`