How do you evaluate the definite integral #int (x^45)(cos(x^46)) dx# from #[0,(pi)^(1/46)]#?
1 Answer
Mar 26, 2016
Explanation:
We have
#int_0^(pi^(1//46))x^45cos(x^46)dx#
Substituting, let
Multiply the interior of the integral by
#=1/46int_0^(pi^(1//46))cos(x^46)*46x^45dx#
Now substitute in for
#"bound of"# #0->" "0^46=0" "larr"new bound"#
#"bound of"# #pi^(1//46)->" "(pi^(1//46))^46=pi" "larr"new bound"#
This gives us the integral of
#=1/46int_0^picos(u)du#
Which then becomes
#=1/46[sin(u)]_0^pi=1/46(sin(pi)-sin(0))=1/46(0-0)=0#