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How to solve this~~ (see attachment soon)
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\[\int\limits_{?}^{?} x \sqrt{4-x^{4}} dx\]
i set u=x\[1/2 \int\limits_{}^{} \sqrt{4-u^{2}} du\]^2 than i got
u =x^2
Do a trig sub, u=sin(theta)
du=cos(theta)dtheta
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Woops, u=2sin(theta). Sorry
You can directly use formula...
wait~ why u=2sin (theta)?
Because then you can factor out 4 from the radical.
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