$\displaystyle \int \frac{7x}{\sqrt{16-x^4}} dx = 7\int \frac{x}{\sqrt{4^2-(x^2)^2}} dx$

The arcsin rule can be used here, right? I'm not sure what to do with the x in the numerator to be able to apply the arcsin rule.

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- Jul 3rd 2008, 10:18 AMdataspotintegration arcsin rule
$\displaystyle \int \frac{7x}{\sqrt{16-x^4}} dx = 7\int \frac{x}{\sqrt{4^2-(x^2)^2}} dx$

The arcsin rule can be used here, right? I'm not sure what to do with the x in the numerator to be able to apply the arcsin rule. - Jul 3rd 2008, 10:31 AMSimplicity
First, you will need to use the substitution of $\displaystyle u = x^2$ to obtain $\displaystyle \frac 72 \int \frac 1{\sqrt{16 - u^2}} \ \mathrm{d}u$.

Then you can apply the arcsin rule which is $\displaystyle \int \frac{1}{\sqrt{a^2 - x^2}} \ \mathrm{d}x = \mathrm{arcsin}\left(\frac{x}{a}\right)$.

Remember to change the $\displaystyle u$ back to $\displaystyle x^2$ in the final answer. - Jul 3rd 2008, 10:39 AMdataspot