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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.

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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.

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    $\begingroup$ This worked for me and makes sense. Any hint as to how one might effectively spot such an approach or is it simply practice? $\endgroup$ – Edward Jan 3 '14 at 17:24
  • $\begingroup$ @Edward Well if you have (x^m)*( a + bx^n)^p, if p is a hole number than make x= t^k , where k is the denominator of p. If (m+1)/n is hole make a+bx^m=t^k and if (m+1)/n + p is hole than make a*x^(-n) + b=t^k. Hope this little trick helps you in problems of this kind.:) $\endgroup$ – randomname Jan 3 '14 at 21:01
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    $\begingroup$ @user109218, what is a hole number? Edward, practice is probably your best bet. What made the substitution work here was the appearance of an odd-degree polynomial in $x$ (namely the $x$ in the denominator) outside of the square root, which could be converted into something of even degree and an $x\,dx$. $\endgroup$ – Barry Cipra Jan 3 '14 at 22:02
  • $\begingroup$ @BarryCipra haha me and my bad english i meant whole number haha $\endgroup$ – randomname Jan 3 '14 at 22:17
  • $\begingroup$ @user109218, in that case, what's the meaning of "where k is the denominator of p"? $\endgroup$ – Barry Cipra Jan 3 '14 at 22:25
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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 $$\int\frac{1}{\sin\theta}d\theta=\bigg |\ln\left(\tan\frac{\theta}{2}\right)\bigg |+C.$$

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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.

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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

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$4\int\frac{\cos^2\theta}{\sin\theta}d\theta=4\int(\frac{1}{\sin\theta}-\sin\theta)d\theta=-4\int\frac{1}{1-\cos^2\theta}d\cos\theta+4 \cos\theta$

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