Step 1: Remove the inverse function by substitution:
Let $x = \sin\theta$. Then $dx = \cos\theta\,d\theta$ and $\sin^{-1}x = \theta$.
The limits change: $x = 0$ gives $\theta = 0$, and $x = 1$ gives $\theta = \dfrac{\pi}{2}$.
Step 2: Rewrite the integral:
Every part of the integrand is now in terms of $\theta$.
\[ I = \int_0^{\pi/2} \theta \sin\theta\cos\theta\,d\theta = \frac{1}{2}\int_0^{\pi/2} \theta \sin 2\theta\,d\theta \]
Here we used $2\sin\theta\cos\theta = \sin 2\theta$.
Step 3: Integrate by parts:
Take $u = \theta$ and $dv = \sin 2\theta\,d\theta$. Then $du = d\theta$ and $v = -\dfrac{\cos 2\theta}{2}$.
\[ \int \theta\sin 2\theta\,d\theta = -\frac{\theta\cos 2\theta}{2} + \frac{1}{2}\int \cos 2\theta\,d\theta = -\frac{\theta\cos 2\theta}{2} + \frac{\sin 2\theta}{4} \]
Step 4: Put in the limits:
At $\theta = \dfrac{\pi}{2}$: $-\dfrac{\pi}{2}\cdot\dfrac{\cos\pi}{2} + \dfrac{\sin\pi}{4} = \dfrac{\pi}{4} + 0$.
At $\theta = 0$: both terms are $0$.
So $\displaystyle\int_0^{\pi/2}\theta\sin 2\theta\,d\theta = \dfrac{\pi}{4}$.
Step 5: Finish:
\[ I = \frac{1}{2}\cdot\frac{\pi}{4} = \frac{\pi}{8} \]
Final Answer:
Both methods agree, and the integral equals $\dfrac{\pi}{8}$. \[ \boxed{\dfrac{\pi}{8}} \]