<s>[INST] How to integrate $\int{\frac{dx}{\sqrt{16-9x^2}}}$ Me again, probably someone is going to blame me. I have this:
$$\int{\frac{dx}{\sqrt{16-9x^2}}}$$
I have asked an old teacher of mine an he said me I should let $x=\frac{4}{3}\sin{\big(\frac{3}{4}x\big)}$ but I don't how he realise that. I have tried $u-substitution$ and Integration by Parts but the square root is making the whole thing much harder.
I tried multiplying both terms by $\sqrt{16-9x^2}$ but nothing. Also tried to factor to eliminate square root but nothing.
I don't want that anyone solve my integral, just give me a hint about how do it, and how I can realise what to do in this cases :)
 [/INST] The given integral is:
$\int{\frac{dx}{\sqrt{16-9x^2}}}$.
This integral can remark as: $\int{\frac{dx}{\sqrt{4^2-(3x)^2}}}$.
By received substition 3x with $4\sin t$ it is $3x=4\sin t$ have:
$3x=4\sin t$ $\Rightarrow$ $x=\frac{4}{3}\sin t$.
Hence we: $dx=\frac{4}{3}\cos t$
If this data replacement the intial integral we have:
$\int{\frac{dx}{\sqrt{16-9x^2}}}$=$\int{\frac{dx}{\sqrt{4^2-(3x)^2}}}$=$\int\frac{4\cos t dt}{3\sqrt{16-(4\sin t)^2}}$=$\frac{4}{3}\int\frac{cos t dt}{\sqrt{16-16\sin^2 t}}$=$\frac{4}{3}\int\frac{cos t dt}{\sqrt{16(1-\sin^2 t)}}$=$\frac{4}{3}\int\frac{cos t dt}{\sqrt{16}\cdot\sqrt{1-\sin^2 t}}$=$\frac{4}{3}\int\frac{cos t dt}{4\cdot\cos t}$=$\frac{4}{3}\cdot\frac{1}{4}\int{dt}$=$\frac{1}{3}t$
By substition of the above have:
$3x=4\sin t$ $\Rightarrow$ $\sin t=\frac{3}{4}\cdot x$ $\Rightarrow$ $ t=\arcsin \frac{3}{4}\cdot x $.
Theres definitly the given integral is:
$\int{\frac{dx}{\sqrt{16-9x^2}}}$=$\frac{1}{3}t$=$\frac{1}{3}\arcsin \frac {3}{4}\cdot x+C $.
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