[tex]\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx[/tex]
Setting [tex]x=5\sin\theta[/tex], you have [tex]\mathrm dx=5\cos\theta\,\mathrm d\theta[/tex]. Then the integral becomes
[tex]\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta[/tex]
[tex]\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta[/tex]
[tex]\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta[/tex]
[tex]\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta[/tex]
Now, [tex]\sqrt{x^2}=|x|[/tex] in general. But since we want our substitution [tex]x=5\sin\theta[/tex] to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means [tex]\theta=\sin^{-1}\dfrac x5[/tex], which implies that [tex]\left|\dfrac x5\right|\le1[/tex], or equivalently that [tex]|\theta|\le\dfrac\pi2[/tex]. Over this domain, [tex]\cos\theta\ge0[/tex], so [tex]\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta[/tex].
Long story short, this allows us to go from
[tex]\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta[/tex]
to
[tex]\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta[/tex]
[tex]\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta[/tex]
Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get
[tex]\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta[/tex]
Then integrate term-by-term to get
[tex]\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)[/tex]
[tex]=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C[/tex]
Now undo the substitution to get the antiderivative back in terms of [tex]x[/tex].
[tex]=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C[/tex]
and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to
[tex]=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C[/tex]
