Question

Integration of sec^2x/[(a/b)^2+tan^2x]?

Answer 1

`I=b/aarctan((btan(x))/a)+C`

Explanation:

We want to solve

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

For simplicity substitute `d=a/b`

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

Make a substitution `u=tan(x)=>(du)/dx=sec^2(x)`

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

Factor out the `d^2`

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

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

Make a substitution `s=u/d=>(ds)/(du)=1/d`

`I=1/dint1/(s^2+1)ds`

`=1/darctan(s)+C`

Now we just need to substitute back

Remember `s=u/d, u=tan(x), d=a/b`

`I=1/darctan(s)+C`

`=1/darctan(u/d)+C`

`=1/darctan(tan(x)/d)+C`

`=1/(a/b)arctan(tan(x)/(a/b))+C`

`=b/aarctan((btan(x))/a)+C`