How to integrate $\int{\frac{dx}{\sqrt{16-9x^2}}}$

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.

Move you mouse over the gray area for the answer.

$$ I = \int \dfrac{dx}{\sqrt{16-9x^2}} = \int \dfrac43 \dfrac{du}{\sqrt{16 - 9 \times \dfrac{16}9 u^2}} = \int \dfrac43 \dfrac{du}{4\sqrt{1-u^2}} = \dfrac13 \arcsin(u) + c = \dfrac13 \arcsin \left( \dfrac34x\right) + c$$

Let $x=\frac{4}{3}u$.${}{}{}{}$ After some algebra, we end up with the familiar integral $$\int\frac{k}{\sqrt{1-u^2}}\,du$$ for a not difficult to compute constant $k$.

Motivation The bottom looks like $\sqrt{1-x^2}$, except that the constants are wrong. One way to think about it is to note that the bottom is equal to $4\sqrt{1-\frac{9x^2}{16}}$, which is closer to the desired shape. Now we would like to have $\frac{9x^2}{16}=u^2$, which is achieved by setting $u=\frac{3x}{4}$. We did it a little more efficiently, by deciding we wanted a $16$ where there is a $9$.

All you need here is to apply the well-known formula:

$$\int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin{\frac{x}{a}}+C.$$

Therefore we get that:

$$\int\frac{dx}{\sqrt{16-9x^2}}=\frac{1}{3}\int\frac{dx}{\sqrt{(\frac{4}{3})^2-x^2}}=\frac{1}{3}\arcsin\left(\frac{3}{4}x\right)+C.$$

NOTE: it's important to recognize the elementary integrals because in some situations when dealing with more complex integrals, we know how to successfully split it into elementary
integrals, when this is possible.

Q.E.D.