## Thomas' Calculus 13th Edition

$\dfrac{1}{6} \sin^{-1}(3t)+\dfrac{t\sqrt {1-9t^2}}{2}+C$
Use formula: $\cos^2 \theta=1-\sin^2 \theta$ and $\cos^2 \theta =(1/2)+(1/2) \cos 2 \theta$ Plug $t =\dfrac{1}{3} \sin \theta \implies dt=\dfrac{1}{3} \cos \theta d \theta$ and $\cos \theta =\sqrt {1-9t^2}$ Then, $\int \sqrt{1-9(\dfrac{1}{9} \sin^2 \theta) } (\dfrac{1}{3} \cos \theta d \theta )=(\dfrac{1}{3}) \int (\cos \theta) (\cos \theta) d\theta =(\dfrac{1}{3}) \int (\dfrac{1}{2}+\dfrac{1}{2} \cos ( 2\theta ))d \theta$ Integrate and substitute in $\theta =\sin^{-1}(3t)$ and $\sin \theta =3t$. $(\dfrac{1}{3})(\dfrac{1}{2} \theta+\dfrac{1}{4} \sin ( 2\theta ))+C=(\dfrac{1}{3})(\dfrac{1}{2} \theta+\dfrac{1}{4} (2 \sin \theta \cos \theta )))+C=(\dfrac{1}{6}) \sin^{-1}(3t)+(3t)(\dfrac{1}{6})(3\sqrt {1-9t^2}+C$ Thus, $\int \sqrt{1-9(\dfrac{1}{9} \sin^2 \theta) } (\dfrac{1}{3} \cos \theta d \theta )=\dfrac{1}{6} \sin^{-1}(3t)+\dfrac{t\sqrt {1-9t^2}}{2}+C$