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 🙂
Best Answer
In general, computing an integral is a guessing process. We usually rely on the fundamental theorem of calculus to provide us the initial guess i.e. if you want to compute $\displaystyle \int f(x) dx$, we need to first guess $F(x)$ such that $F'(x) = f(x)$. Then we make use of the fundamental theorem of calculus to conclude that $\displaystyle \int f(x) dx = F(x) + c$.
If you have an integral of the form $\displaystyle \int \dfrac{dx}{\sqrt{b^2 - a^2x^2}}$, recall that you would seen something similar before. You would have learnt that the derivative of $\arcsin(x)$ is $\dfrac1{\sqrt{1-x^2}}$. But now you the $a$ and $b$ hanging around. The goal now is to convert $\sqrt{b^2 - a^2x^2}$ into something like $\sqrt{1-u^2}$, so that we can then recognize that to be the derivative of $\arcsin(u)$.
Consider $\sqrt{b^2 - a^2x^2}$. The first thing is to pull out the $b$. This gives us $$\sqrt{b^2 - a^2x^2} = b \sqrt{1 - \dfrac{a^2}{b^2}x^2}$$ Now it looks very similar to a scaled version of $\sqrt{1-u^2}$ except for the coefficient in front of $x^2$. This gives us the motivation to make the substitution. $u^2 = \dfrac{a^2}{b^2}x^2$ i.e. $u = \dfrac{a}b x$.
In our case $a = 3$ and $b=4$. Hence, substitute $u = \dfrac34x$. We then get $du = \dfrac34 dx$. Plug this into the integral and recall that the derivative of $\arcsin(u)$ is $\dfrac1{\sqrt{1-u^2}}$ to get the integral.
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