Yes that is the correct integral. But how do you solve it with the tanx = t substitution???

\(\displaystyle t = \tan{x}\)

\(\displaystyle x = \arctan{t}\)

\(\displaystyle dx = \frac{dt}{1+t^2} \)

\(\displaystyle \cos^2{x} = \frac{1}{t^2+1}\)

\(\displaystyle \sin^2{x} = \frac{t^2}{t^2+1}\)

\(\displaystyle \int_0^{\frac{\pi}{4}} \frac{dx}{9\cos^2{x} - \sin^2{x}}

\)

substitute ...

\(\displaystyle \int_0^1 \frac{dt}{9 - t^2}\)

use partial fractions to finish