#int sqrt(9-5x^2) dx# ?

I know that x=asintheta. But idk how to factor out so that i wont have #sqrt5#

1 Answer
Feb 19, 2017

The answer is #=9/(2sqrt5)arcsin(sqrt5/3x)+1/2xsqrt(9-5x^2)+C#

Explanation:

We perform this integral by substitution

Let #x=3/sqrt5sinu#

#dx/(du)=3/sqrt5cosu#

#dx=3(du)/sqrt5cosu#

and

#sqrt(9-5x^2)=sqrt(9-5*9/5sin^2u)#

#=3sqrt(1-sin^2u)#

#=3cosu#

So,

#intsqrt(9-5x^2)dx=int3/sqrt5cosu*3cosudu#

#=9/sqrt5intcos^2udu#

#cos2u=2cos^2u-1#

#cos^2u=(1+cos(2u))/2#

So,

#intsqrt(9-5x^2)dx=9/sqrt5int(1+cos(2u))/2(du)#

#=9/(2sqrt5)int(1+cos(2u))du#

#=9/(2sqrt5)(u+sin(2u)/2)#

#=9/(2sqrt5)(arcsin(sqrt5/3x)+sinucosu)#

#=9/(2sqrt5)(arcsin(sqrt5/3x)+sqrt5/3xsqrt(9-5x^2)/3)+C#

#=9/(2sqrt5)arcsin(sqrt5/3x)+1/2xsqrt(9-5x^2)+C#