University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{1}{6} \sin^{-1}(3t)+\dfrac{t\sqrt {1-9t^2}}{2}+C$
Let us consider $t =\dfrac{1}{3} \sin \theta \implies dt=\dfrac{1}{3} \cos \theta d \theta$ and $\cos \theta =\sqrt {1-9t^2}$ Now, the given integral becomes: $\int \sqrt{1-9(\dfrac{1}{9} \sin^2 \theta) } (\dfrac{1}{3} \cos \theta d \theta )=\dfrac{1}{3} \int (\cos \theta) (\cos \theta) d\theta$ As we know that $\cos^2 \theta=1-\sin^2 \theta$ and $\cos^2 \theta =(1/2)+(1/2) \cos 2 \theta$ $= \dfrac{1}{3} \int (\dfrac{1}{2}+\dfrac{1}{2} \cos ( 2\theta ))d \theta$ Now, integrate and plug in $\theta =\sin^{-1}(3t)$ and $t =\dfrac{1}{3} \sin \theta \implies \sin \theta =3t$ we have $\dfrac{1}{3}(\dfrac{1}{2} \theta+\dfrac{1}{4} \sin ( 2\theta ))+C=\dfrac{1}{3}(\dfrac{1}{2} \theta+\dfrac{1}{4} (2 \sin \theta \cos \theta )))+C=\dfrac{1}{6} \sin^{-1}(3t)+\dfrac{1}{6}(3 t ) \sqrt {1-9t^2}+C$ or, $=\dfrac{1}{6} \sin^{-1}(3t)+\dfrac{t\sqrt {1-9t^2}}{2}+C$