Question

Integral of dx/a+b*tan^2 x =?

Answer 1

`1/((a/b-1)b)(x+sqrt(b/a)arctan(tanxsqrt(b/a))) + C`

Explanation:

`int(1/(a+btan^2x))dx`
put `tanx=t` `=>sec^2x dx=dt` `sec^2x=tan^2x+1`
`=>sec^2x=t^2+1`
`=>dx=dt/(t^2+1)`
`int(1/(a+bt^2))dt/(t^2+1)`
`int(1/(b(a/b+t^2)))dt/(t^2+1)`
multiply numerator and denominator with `a/b-1` we get
`int((a/b-1)/((a/b-1)(b(a/b+t^2)))dt/(t^2+1)`

add and subtract `t^2` in numerator we get
`int((t^2+a/b-1-t^2)/((a/b-1)(b(a/b+t^2)))dt/(t^2+1)`
`1/((a/b-1)b)int((t^2+a/b-1-t^2)/((a/b+t^2)(t^2+1)))dt`
`1/((a/b-1)b)int((-1)/((a/b+t^2))+1/(t^2+1)))dt`
`1/((a/b-1)b)(arctan(t)+sqrt(b/a)arctan(tsqrt(b/a)) + C`
substituting `t=tanx`
`1/((a/b-1)b)(arctan(tanx)+sqrt(b/a)arctan(tanxsqrt(b/a)) + C`
`arctan(tanx)=x`
`1/((a/b-1)b)(x+sqrt(b/a)arctan(tanxsqrt(b/a))) + C`