[25sin^2(x)+9cos^2(x)]=[9+16sin^2(x)] Why?

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I'm doing a problem where I need to find the maximum curvature. I'm at a point where I need to simplify the denominator and reduce the funcitions in the bottom. I don't understand how to simplify:

[25sin^2(x)+9cos^2(x)]

To:

[9+16sin^2(x)]

What is getting factored out? Or is this completing the square somehow?

Thanks for any help.
 
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the simplyfication is based on the fact [ sin^2(x) + cos^2(x) = 1 ]

write down ur equation as: [ (16+9)sin^2(x) + 9cos^2(x) ] ====
[ 16sin^2(x) + 9sin^2(x) + 9cos^2(x) ] =====

[ 16sin^2(x) + 9{ sin^2(x) + cos^2(x) } ] ====

[ 16sin^2(x) + 9(1) ] ====

[ 16 sin^2(x) + 9 ]
 
abluphoton said:
the simplyfication is based on the fact [ sin^2(x) + cos^2(x) = 1 ]

write down ur equation as: [ (16+9)sin^2(x) + 9cos^2(x) ] ====
[ 16sin^2(x) + 9sin^2(x) + 9cos^2(x) ] =====

[ 16sin^2(x) + 9{ sin^2(x) + cos^2(x) } ] ====

[ 16sin^2(x) + 9(1) ] ====

[ 16 sin^2(x) + 9 ]

Yeah I'm retarded. I knew it would be something that simple. Thanks
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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