int sqrt(9-5x^2) dx95x2dx ?

I know that x=asintheta. But idk how to factor out so that i wont have sqrt55

1 Answer
Feb 19, 2017

The answer is =9/(2sqrt5)arcsin(sqrt5/3x)+1/2xsqrt(9-5x^2)+C=925arcsin(53x)+12x95x2+C

Explanation:

We perform this integral by substitution

Let x=3/sqrt5sinux=35sinu

dx/(du)=3/sqrt5cosudxdu=35cosu

dx=3(du)/sqrt5cosudx=3du5cosu

and

sqrt(9-5x^2)=sqrt(9-5*9/5sin^2u)95x2=9595sin2u

=3sqrt(1-sin^2u)=31sin2u

=3cosu=3cosu

So,

intsqrt(9-5x^2)dx=int3/sqrt5cosu*3cosudu95x2dx=35cosu3cosudu

=9/sqrt5intcos^2udu=95cos2udu

cos2u=2cos^2u-1cos2u=2cos2u1

cos^2u=(1+cos(2u))/2cos2u=1+cos(2u)2

So,

intsqrt(9-5x^2)dx=9/sqrt5int(1+cos(2u))/2(du)95x2dx=951+cos(2u)2(du)

=9/(2sqrt5)int(1+cos(2u))du=925(1+cos(2u))du

=9/(2sqrt5)(u+sin(2u)/2)=925(u+sin(2u)2)

=9/(2sqrt5)(arcsin(sqrt5/3x)+sinucosu)=925(arcsin(53x)+sinucosu)

=9/(2sqrt5)(arcsin(sqrt5/3x)+sqrt5/3xsqrt(9-5x^2)/3)+C=925(arcsin(53x)+53x95x23)+C

=9/(2sqrt5)arcsin(sqrt5/3x)+1/2xsqrt(9-5x^2)+C=925arcsin(53x)+12x95x2+C