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Hint says you may like to use the identity sin(theta)cos(theta)= sin(2theta)/2

However, I think I found a way to use 1-sin^2(theta)=cos^2(theta)

First, (2x)^4 = 16x^4

So make it 16(1-x^2)^2.

Take the 16 out of the root and the integral and you have:

[tex]4\int\sqrt{1-(x^{2})^{2}}xdx[/tex]

Sub in sin(theta) for x^2. Sub in cos^2(theta) for 1-sin^2(theta). Sub in [tex]\sqrt{sin(\vartheta)[/tex] for the x outside the root and [tex]\frac{1}{2\sqrt{sin(\vartheta)}}[/tex] for dx.

Cancel out the sqrt{sin} on the top and bottom and drag the 1/2 outside the integral. Anti-derivative of cos is sin and you have 2sin(theta). Sin(theta) = x^2.

And so the final answer is; [tex]2x^{2}+c[/tex]

Correct?