We perform this integral by substitution
Let x=3/sqrt5sinux=3√5sinu
dx/(du)=3/sqrt5cosudxdu=3√5cosu
dx=3(du)/sqrt5cosudx=3du√5cosu
and
sqrt(9-5x^2)=sqrt(9-5*9/5sin^2u)√9−5x2=√9−5⋅95sin2u
=3sqrt(1-sin^2u)=3√1−sin2u
=3cosu=3cosu
So,
intsqrt(9-5x^2)dx=int3/sqrt5cosu*3cosudu∫√9−5x2dx=∫3√5cosu⋅3cosudu
=9/sqrt5intcos^2udu=9√5∫cos2udu
cos2u=2cos^2u-1cos2u=2cos2u−1
cos^2u=(1+cos(2u))/2cos2u=1+cos(2u)2
So,
intsqrt(9-5x^2)dx=9/sqrt5int(1+cos(2u))/2(du)∫√9−5x2dx=9√5∫1+cos(2u)2(du)
=9/(2sqrt5)int(1+cos(2u))du=92√5∫(1+cos(2u))du
=9/(2sqrt5)(u+sin(2u)/2)=92√5(u+sin(2u)2)
=9/(2sqrt5)(arcsin(sqrt5/3x)+sinucosu)=92√5(arcsin(√53x)+sinucosu)
=9/(2sqrt5)(arcsin(sqrt5/3x)+sqrt5/3xsqrt(9-5x^2)/3)+C=92√5(arcsin(√53x)+√53x√9−5x23)+C
=9/(2sqrt5)arcsin(sqrt5/3x)+1/2xsqrt(9-5x^2)+C=92√5arcsin(√53x)+12x√9−5x2+C