Integral $\int \frac{\sqrt{16-x^2}}{x} \mathrm{d}x$ The problem is $\displaystyle\int\frac{\sqrt{16-x^2}}{x}\mathrm{d}x$.  I've attempted to use a trig substitution with $x=4\sin\theta$ and $\mathrm{d}x=4\cos\theta\ \mathrm{d}\theta$.  This yields  $ \displaystyle 4 \int\frac{\cos^2\theta}{\sin\theta}\mathrm{d}\theta$ and I attempted to substitute $1-\sin^2 \theta$ for the numerator but that did not appear to yield a tractable integral either. (Similar result attempting to substitute a double angle formula.) I attempted to do an integration by parts with $\displaystyle 4\int\frac{\cos\theta}{\sin\theta}\cos\theta\ \mathrm{d}\theta$ and $u=\cos\theta$ and $\displaystyle \mathrm{d}v=\frac{\cos\theta}{\sin\theta}\mathrm{d}\theta$ which gets me $\displaystyle \cos\theta\ln\sin\theta + \int\ln(\sin\theta) \sin\theta\ \mathrm{d} \theta$ and I don't know how to solve that integral either. 
 A: Write the integral as
$$\int{\sqrt{16-x^2}\over x^2}x\,dx$$
then let $u^2=16-x^2$, so that $u\,du=-x\,dx$ and the substitution gives
$$-\int{u\over16-u^2}u\,du=\int\left(1-{16\over16-u^2}\right)\,du$$
Partial fractions should finish things off.
A: Another way.
After your getting
$$4\int\frac{\cos^2\theta}{\sin\theta}d\theta=4\int\frac{1-\sin^2\theta}{\sin\theta}d\theta=4\int\left(\frac{1}{\sin\theta}-\sin\theta\right)d\theta,$$
you can use
$$\int\frac{1}{\sin\theta}d\theta=\ln\bigg|\tan\frac{\theta}{2}\bigg |+C.$$
A: To Complete your work:
Let $u=\cos \theta$ then $du=-\sin\theta d\theta=-(1-\cos^2\theta)\frac{d\theta}{\sin\theta}$ so $\frac{du}{u^2-1}=\frac{d\theta}{\sin\theta}$ hence
$$\int\frac{\cos^2\theta d\theta}{\sin\theta}=\int\frac{u^2du}{u^2-1}$$
and now the rest is simple.
A: Hint : 
First use  $x^2=t\implies 2xdx=dt$ to reduce  $\int\frac{\sqrt{16-t}}{t}dt$;
Now substitute $16-t=z^2 \implies dt = -2zdz$ which reduces it to  $\int\frac{-2z^2}{16-z^2}dz$. Now use partial fractions 
A: $4\int\frac{\cos^2\theta}{\sin\theta}d\theta=4\int(\frac{1}{\sin\theta}-\sin\theta)d\theta=-4\int\frac{1}{1-\cos^2\theta}d\cos\theta+4 \cos\theta$
