Trigonometric substitution. Pretty confused where constant comes from. (fixed)

Find ∫from .6 to 0 x^2/ sqrt(9-25x^2) dx

My teacher worked this on the board a little confused
O obviously the trig sub is asintheta. But it isn't in the right form yet. So get it there you pull out a 25 --) sqrt(25(9/25) - x^2 ) 5sqrt((9/25)-x^2 so x= (3/5) sintheta so dx = 3/5costheta. so you change the bounds pi/2 to 0
When you replace everything you get
∫pi/2 to 0 (9/25) times ( sin^2θ/ 3cosθ) times (3/5) cosdθ

Where does that (9/25) come from?

I understand how to do the rest, just totally confused where that constant comes

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Dick
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Find ∫from .6 to 0 x^2/ sqrt(9-25x^2) dx

My teacher worked this on the board a little confused
O obviously the trig sub is asintheta. But it isn't in the right form yet. So get it there you pull out a 25 --) sqrt(25(9/25) - x^2 ) 5sqrt((9/25)-x^2 so x= (3/5) sintheta so dx = 3/5costheta. so you change the bounds pi/2 to 0
When you replace everything you get
∫pi/2 to 0 (9/25) times ( sin^2θ/ 3cosθ) times (3/5) cosdθ
^
||
Where does that (9/25) come from?

I understand how to do the rest, just totally confused where that constant comes
If x=(3/5)sin(theta) then the x^2 in the numerator becomes (9/25)sin(theta)^2. Is that the factor you are looking for?

If x=(3/5)sin(theta) then the x^2 in the numerator becomes (9/25)sin(theta)^2. Is that the factor you are looking for?
Thanks, it is that factor. But if you think about the cos too. It would be (3/5sinθ)2 --) 9/25 sintheta on the numerator and 3/5 cos theta on the demoniator. So wouldn't that give you 25/9 times 5 if you keep the 3 where she kept it with the cos.

Dick