Trigonometric substitution for $\int\frac1{x^2\sqrt{16-x^2}}\ dx$

Question

The original problem is $$\int\frac1{x^2\sqrt{16-x^2}}\ dx$$ I went all the way (using $x=4\sin\theta$) to $$\int\frac1{16\sin^2\theta\cos\theta}\ d\theta$$ How would I continue this, since the sines and cosines are in the denominator? I thought of secants and cosecants, but I think that's wrong.

Answer 1

Let $x=4\sin t$, $\quad -\pi/2 \le t\le\pi/2,\quad$then we get \begin{align*} \int\frac{dx}{x^2\sqrt{16-x^2}}&=\int\frac{4\cos t\,dt}{16\sin^2 t\;(4\cos t)}\\[5pt] &=\frac1{16}\int \csc^2 t\,dt\\[5pt] &=-\frac1{16}\cot t+C\\[3pt] &=-\frac1{16}\frac{\sqrt{1-(x/4)^2}}{x/4}+C\\[3pt] &=-\frac{\sqrt{16-x^2}}{16x}+C \end{align*}

Answer 2

Let $x=4\sin\theta$. Therefore, $dx=4\cos\theta d\theta$.

From there, you get $$\int\frac{4\cos\theta d\theta}{16\sin^2\theta*4\cos\theta}$$ This leaves you with $$\int\frac{d\theta}{16\sin^2\theta}=\frac1{16}\int\csc^2\theta$$

Answer 3

Let $u=1/x$ , $du=-dx/x^2$, Then

$$\int\frac1{x^2\sqrt{16-x^2}}\ dx$$

=$$\int\frac{-1}{\sqrt{16-\frac{1}u^2}}\ dx$$

=$$\int\frac{-u}{\sqrt{16u^2-1}}\ dx$$

Then it is easy to solve.