Answer is: option1
\( \displaystyle \int_{0}^{\pi} \sqrt{\cos^2 t + 1}\,dt \)Solution:
The parametric form for the length of arc is \[ L = \int_{0}^{\pi} \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}} \, dt. \]
\( x = \sin t \)
\( y = t \)
\[ \frac{dx}{dt} = \cos t, \quad \frac{dy}{dt} = 1 \]
Then \[ L = \int_{0}^{\pi} \sqrt{(\cos t)^{2} + 1^{2}} \, dt = \int_{0}^{\pi} \sqrt{\cos^{2} t + 1} \, dt. \]
