(a) The substitution gives \[dx=3\sec^2\theta\,d\theta,\qquad9+x^2=9(1+\tan^2\theta)=9\sec^2\theta.\] At \(x=0\), \(\theta=0\); at \(x=3\), \(\theta=\pi/4\). Hence \[I=\int_0^{\pi/4}\frac{81\sec^2\theta}{81\sec^4\theta}\,d\theta=\int_0^{\pi/4}\cos^2\theta\,d\theta.\]

(b) Use \(\cos^2\theta=(1+\cos2\theta)/2\): \[I=\left[\frac\theta2+\frac14\sin2\theta\right]_0^{\pi/4}=\boxed{\frac\pi8+\frac14=\frac{\pi+2}8}.\]