Question

how do you find the integral of 1/sqrt(9-x^2)

Answer 1

u have formula for dis look into ur text book

Answer 2

dx/sqre root(a^2-x^2)

Answer 3

consider x = 3 sin(t)

dx = 3 cos(t) dt

Answer 4

\[\int\frac{1}{\sqrt{9-x^2}}dx\]
\[x=3~sin(t)~:~sin^{-1}(\frac x3)=t\\
dx=3~cos(t)~dt\]
\[\int\frac{3~cos(t)}{\sqrt{9-(3~sin(t))^2}}dt\]
\[\int\frac{3~cos(t)}{\sqrt{9-9~sin^2(t)}}dt\]
\[\int\frac{3~cos(t)}{3\sqrt{1-sin^2(t)}}dt\]
\[\int\frac{3~cos(t)}{3~cos(t)}dt\]
\[\int~dt=t+C\]
\[sin^{-1}(\frac x3)+C\]