Question

设I=∫﹙a+x）／根号下（a²-x²）dx 则I=?求详解

Answer 1

I=∫﹙a+x）／根号下（a²-x²）dx 令x=asint,则dx=acostdt

=∫﹙a+asint）／(acost)*acostdt

=∫﹙a+asint）dt

=at-acost+C

=a*arcsin(x/a)-√(a²-x²) +C