Show that the substitution $t=\tan\theta$ transforms the integral ${\int}\frac{d\theta}{9\cos^2\theta+\sin^2\theta}$, into ${\int}\frac{dt}{9+t^2}$

To begin with the $d\theta$ on the top of the fraction threw me off but I'm assuming it's just another way of representing:

$${\int}\frac{1}{9\cos^2\theta+\sin^2\theta}\,d\theta$$

I tried working backwards

$$\frac{d}{d\theta}\tan\theta=\sec^2\theta\,\,\,\,{\Rightarrow}\,\,\,\,d\,\tan\theta=\sec^2\theta\,d\theta$$ $${\Rightarrow}\,{\int}\frac{\sec^2\theta\,d\theta}{9+\tan^2\theta}$$

$$\tan\theta=\frac{\sin\theta}{\cos\theta}\,\,\,\,{\Rightarrow}\,\,\,\,\tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}$$

$${\Rightarrow}\,{\int}\frac{\sec^2\theta\,d\theta}{\left(9+\dfrac{\sin^2\theta}{\cos^2\theta}\right)}$$

$$9=\frac{9\cos^2\theta}{\cos^2\theta}\,\,\,\,{\Rightarrow}\,\,\,\,{\int}\frac{\sec^2\theta\,d\theta}{\left(\dfrac{9\cos^2\theta}{\cos^2\theta}+\dfrac{\sin^2\theta}{\cos^2\theta}\right)}\,\,\,\,{\Rightarrow}\,\,\,\,{\int}\frac{\sec^2\theta\,d\theta}{\left(\dfrac{9\cos^2\theta+\sin^2\theta}{\cos^2\theta}\right)}$$

$${\Rightarrow}\,\,\,\,{\int}\frac{\color{red}{\cos^2\theta\,\sec^2\theta}\,d\theta}{9\cos^2\theta+\sin^2\theta}$$

Now I have to prove that $$\cos^2\theta\,\sec^2\theta=1$$ but I don't think it is... What have I done wrong? Regards Tom