Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

1.33 Integral associated with the circle

(ii) For the circle, we substitute $x=a\sin u$ in

\int_{0}^{t}(a^{2}-x^{2})^{1/2}\,dx.

When $x=t$ , we have: $u=\sin^{-1}(\frac{t}{a})$ . Thus

\int_{0}^{t}(a^{2}-x^{2})^{\frac{1}{2}}\,dx=\,{\int_{0}^{\sin^{-1}(\frac{t}{a}% )}(a^{2}-a^{2}\sin^{2}u)^{\frac{1}{2}}.a\cos u\,du}