Question

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

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

Answer 1

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`