Question

How do you find the integral of `(6t)/sqrt(36-t^2)`?

Answer 1

Consider 36 -`t^2` =u, so that tdt=`-1/2` du

`int (6tdt)/sqrt(36-t^2)=int -6/2 (du)/sqrtu` = -3*2 `u^(1/2)`

= -6`sqrt(36-t^2)` +C